Physics
52 Solved Past Papers
2008 to 2024 Archives
Electromagnetism Past Papers
Solved past paper MCQs for Electromagnetism from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Official printable A4 booklet containing all 52 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
For a positive charged particle (q) moving with a velocity (v) in a magnetic Field of flux density B, the force (F) acting on the charge particle is given by the expression?
The macroscopic magnetic force (Lorentz force) acting on a moving point charge is defined fundamentally by the vector cross product of velocity and the magnetic field.
The force \( \mathbf{F} \) is a vector that is always strictly perpendicular to both the velocity vector \( \mathbf{v} \) and the magnetic field vector \( \mathbf{B} \).
In mathematics, a vector that is perpendicular to two other vectors is found using the cross product.
The magnitude is scaled by the scalar value of the electric charge \( q \).
Therefore, the correct standard expression is \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \).
Why other options are incorrect:
Option C incorrectly places the charge in the denominator. Options B and D mathematically misarrange the vector relations and isolate the scalar charge \( q \) using invalid vector division.
Magnetic flux density is a measure of the concentration of magnetic field lines within a given area. It is synonymous with the magnetic field vector \( \mathbf{B} \).
Formula:
$$ B = \frac{\Phi}{A} $$
Solution:
Look at the formula: \( B \) is equal to magnetic flux (\( \Phi \)) divided by Area (\( A \)).
The SI unit for magnetic flux is the Weber (Wb).
The SI unit for area is square meters (\( \text{m}^2 \)).
Dividing them gives \( \frac{\text{Wb}}{\text{m}^2} \), which is expressed mathematically with a negative exponent as \( \text{Wb m}^{-2} \) (equivalent to 1 Tesla).
Why other options are incorrect:
Option C is the unit for total flux, not density. Options A and D represent incorrect dimensional scaling by length rather than area.
The formula \( \Phi = \mathbf{B} \cdot \mathbf{A} \) represents
A
Magnetic flux
B
Electric flux
C
Electric flux density
D
Gravitational flux
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In physics, "flux" generally refers to the flow or permeation of a vector field across a defined surface area. The specific type of flux depends entirely on the specific vector field being measured.
Formula:
$$ \Phi_B = \mathbf{B} \cdot \mathbf{A} $$
Solution:
Identify the primary vector in the equation: \( \mathbf{B} \) is the universal symbol for the Magnetic Field (or magnetic induction).
Identify the operation: The dot product with the area vector \( \mathbf{A} \) calculates the projection of this field through the surface.
Because the field is magnetic, the resulting scalar value is explicitly defined as the Magnetic flux.
Why other options are incorrect:
Option B (Electric flux) uses the electric field vector \( \mathbf{E} \) (i.e., \( \Phi_E = \mathbf{E} \cdot \mathbf{A} \)). Option D uses the gravitational field vector \( \mathbf{g} \). Option C is a density, not a total flux calculation.
The wire of length \( 100 \ \text{cm} \) is perpendicular to the magnetic field of \( 0.5 \ \text{T} \). If it carries \( 10 \ \text{A} \) of current then force acting on the wire will be:
A
\( 25 \ \text{N} \)
B
\( 10 \ \text{N} \)
C
\( 5 \ \text{N} \)
D
\( 50 \ \text{N} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A straight wire carrying an electric current through a uniform external magnetic field experiences a macroscopic Lorentz force.
Formula:
$$ F = ILB \sin(\theta) $$
Solution:
First, convert all units to standard SI units: The length \( L = 100 \ \text{cm} = 1.0 \ \text{m} \).
Identify the other givens: Magnetic field \( B = 0.5 \ \text{T} \), Current \( I = 10 \ \text{A} \).
Determine the angle: The problem states the wire is "perpendicular" to the field, so \( \theta = 90^\circ \), and \( \sin(90^\circ) = 1 \).
Calculate the force: \( F = (10)(1.0)(0.5)(1) \).
\( F = 5 \ \text{N} \).
Why other options are incorrect:
Option D occurs if one forgets to convert centimeters to meters, resulting in \( 10 \times 100 \times 0.5 = 500 \) or similar scaling errors. Options A and B are simple math errors.
If \( 0.5 \ \text{T} \) field is applied over area of 2-meter square which lies at an angle of 60 degree with the field, then the resulting flux will be:
A
\( 0.5 \ \text{T} \)
B
\( 0.5 \ \text{Wb} \)
C
\( 0.25 \ \text{Wb} \)
D
\( 0.25 \ \text{T} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Magnetic flux is the dot product of the magnetic field vector and the area vector. Care must be taken depending on whether the angle is given relative to the surface plane or the normal vector. Historical Note: In many past papers from this region, "angle with the field" is interpreted as the angle with the area vector unless specified as the plane.
Formula:
$$ \Phi = BA \cos(\theta) $$
Solution:
Identify parameters: \( B = 0.5 \ \text{T} \), \( A = 2 \ \text{m}^2 \).
Use the angle provided exactly as \( \theta = 60^\circ \) .
Option A has the incorrect unit (Tesla is for field, not flux). Options C and D calculate flux using a sine or a \( 30^\circ \) angle, or just have incorrect math/units.
The magnitude of magnetic force will be maximum on current carrying conductor in uniform magnetic field if conductor is placed?
A
Parallel to magnetic field
B
At 45 degree in magnetic field
C
Perpendicular to magnetic field
D
Antiparallel in magnetic field
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A macroscopic current-carrying wire experiences a magnetic Lorentz force. The magnitude relies heavily on its geometric orientation inside the external magnetic field.
Formula:
$$ F = ILB \sin(\theta) $$
Solution:
To maximize the force \( F \), the trigonometric multiplier \( \sin(\theta) \) must be at its maximum value of 1.
This occurs when the angle \( \theta \) between the wire and the field is \( 90^\circ \).
An angle of \( 90^\circ \) corresponds geometrically to being placed strictly perpendicular to the magnetic field lines.
Why other options are incorrect:
Options A and D result in a force of absolutely zero. Option B results in a force reduced by a factor of \( 0.707 \) (\( \sin 45^\circ \)).
Magnetic flux defines the total magnetic field cutting through a defined geometrical area. It has a specific derived SI unit.
Formula:
$$ \Phi = \int \mathbf{B} \cdot d\mathbf{A} $$
Solution:
The SI unit for the magnetic field \( B \) is Tesla (T).
The SI unit for area \( A \) is square meters (\( \text{m}^2 \)).
The product is \( \text{T} \cdot \text{m}^2 \).
In the International System of Units, this combination is given the dedicated name Weber (Wb).
Why other options are incorrect:
Options B, C, and D inappropriately attach extra distance dimensions (meters) to the Weber, creating nonsensical units that do not represent physical flux.
In physics, an area is treated not just as a size (scalar), but as a mathematical vector whose direction is strictly perpendicular (normal) to the physical surface plane.
Formula:
$$ \Phi_B = \mathbf{B} \cdot \mathbf{A} $$
Solution:
Magnetic flux is defined as the measure of magnetic field lines passing through a surface.
Mathematically, this requires a projection of the magnetic field vector onto the normal of the surface.
This projection is calculated using a mathematical dot product between the Magnetic field vector (\( \mathbf{B} \)) and the vector area (\( \mathbf{A} \)).
Why other options are incorrect:
Option A is impossible because you cannot take a dot product with a scalar. Options C and D define flux density/gradients, not the flux itself, and are mathematically structured backward.
The dimension of magnetic field is same as that of:
A
Magnetic flux density
B
Magnetic flux
C
Magnetic force
D
Work done
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In electromagnetism, the vector \( \mathbf{B} \) is commonly referred to by two different but synonymous names depending on context, yet they describe the exact same physical property and share the same dimensions.
Formula:
$$ \mathbf{B} = \frac{\Phi}{A} $$
Solution:
The term "Magnetic Field" or "Magnetic Induction" refers to the vector \( \mathbf{B} \) measured in Tesla.
Because flux \( \Phi \) is field multiplied by area, rearranging gives \( B = \Phi / A \).
Therefore, \( \mathbf{B} \) is mathematically the amount of magnetic flux per unit area.
Thus, its formal descriptive name is Magnetic flux density, proving they have identical dimensions.
Why other options are incorrect:
Option B (Flux) differs by a dimension of area (\( L^2 \)). Option C is a force (Newtons). Option D is energy (Joules). None share the dimensions of Tesla.
Which of the following defines the change in magnetic flux per unit area?
A
Magnetic flux
B
Magnetic force
C
Magnetic dipole
D
Magnetic flux density
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The quantity of magnetic field lines packed into a specific area defines the strength of the field in that region.
Formula:
$$ B = \frac{\Phi}{A} $$
Solution:
The phrase "magnetic flux per unit area" is a direct verbal translation of the formula \( \Phi / A \).
By definition, flux divided by area yields the magnetic field \( \mathbf{B} \).
Another formal scientific term for \( \mathbf{B} \) is Magnetic flux density.
Why other options are incorrect:
Option A is the numerator itself. Option B relates to the physical push/pull on moving charges. Option C refers to a source of magnetic field (like a bar magnet), not the field distribution itself.
What is the dot product of magnetic induction and unit area?
A
Magnetic flux
B
Magnetic induction
C
Magnetic field
D
Magnetic pole
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
This is a direct translation of the fundamental mathematical definition of a physical quantity in electromagnetism.
Formula:
$$ \Phi = \mathbf{B} \cdot \mathbf{A} $$
Solution:
"Magnetic induction" is another term for the magnetic field vector \( \mathbf{B} \).
The text specifies the operation is a "dot product" (\( \cdot \)) with the "area" vector \( \mathbf{A} \).
The mathematical evaluation of \( \mathbf{B} \cdot \mathbf{A} \) generates a scalar quantity known globally as Magnetic flux.
Why other options are incorrect:
Options B and C are the same thing (the vector \( \mathbf{B} \) itself). Option D is a theoretical magnetic monopole or the physical ends of a magnet, completely unrelated to this calculation.
Magnetic flux is a conceptual and mathematical tool used to quantify the overall amount of a magnetic field permeating a given 2D surface.
Solution:
Imagine the magnetic field as a stream of water, and a wire loop as a net.
The flux is the total volume of water passing through the net.
Visually, in physics, this is represented by counting how many imaginary field lines pierce through the specified area.
Therefore, flux is an abstract measure of the number of magnetic lines of force crossing a boundary.
Why other options are incorrect:
Option A is purely geometrical space. Option B defines the density of lines, not the total count. Option C is a time-derivative completely unrelated to static flux.
If a charged particle enters the magnetic field parallel, it will:
A
Deflect toward north
B
Deflect toward south
C
Move straight
D
Move in circular path
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The magnetic force requires a cross product. If vectors are perfectly parallel, their cross product, and therefore the resulting force, collapses to zero.
The dimension of magnetic field strength is same as that of:
A
Magnetic flux
B
Magnetic induction
C
Work done
D
Magnetic force
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In foundational physics terminology, several names describe the exact same underlying vector field denoted by \( \mathbf{B} \).
Solution:
The vector \( \mathbf{B} \) is heavily referenced in physics as either the "magnetic field", "magnetic flux density", or "magnetic induction".
Because "magnetic field strength" (when used colloquially to mean \( \mathbf{B} \)) refers to the exact same Tesla-measured phenomenon, it shares identical dimensions with Magnetic induction.
Why other options are incorrect:
Option A (Flux) is measured in Webers (\( \text{T} \cdot \text{m}^2 \)). Option C is Joules. Option D is Newtons. None of these share the unit of Tesla (\( \text{kg s}^{-2} \text{A}^{-1} \)).
Velocity selector will select only those charge particle whose velocity is given by:
A
\( V = E/B \)
B
\( V = B/E \)
C
\( V = E \times B \)
D
\( V = 0 \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A velocity selector utilizes perpendicular electric and magnetic fields to filter charged particles based on speed.
Formula:
$$ F_e = F_b $$
Solution:
The electric force (\( F_e = qE \)) and magnetic force (\( F_b = qvB \)) are arranged to oppose each other.
For a particle to pass through undeflected, the forces must balance precisely: \( qE = qvB \).
Canceling charge (\( q \)) and solving for velocity gives: \( v = \frac{E}{B} \).
Why other options are incorrect:
Option B is the inverted ratio. Option C is mathematically incorrect dimensionally. Option D implies stationary particles, which cannot pass through the apparatus.
Which surface has greater magnetic flux in same magnetic field, each has an area \( 1\text{m}^2 \)?
A
Circular
B
Rectangular
C
Square
D
Flux is independent of shape
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Magnetic flux depends only on the area's magnitude, the magnetic field strength, and their relative angle, not the geometrical perimeter or shape.
Formula:
$$ \Phi = B \cdot A $$
Solution:
The formula for flux \( \Phi = BA \cos(\theta) \) requires only the numerical value of the Area \( A \).
Whether the \( 1 \ \text{m}^2 \) area is cut into a circle, a long rectangle, or a square, the total number of field lines intersecting that boundary remains identically the same (assuming uniform field and orientation).
Therefore, the flux is entirely independent of the surface's 2D shape.
Why other options are incorrect:
Options A, B, and C incorrectly imply that geometry affects the counting of field lines over a flat plane, which is topologically false.
The force exerted on charge particle will be maximum when it enters the magnetic field at:
A
\( 60^\circ \)
B
\( 90^\circ \)
C
\( 0^\circ \)
D
\( 45^\circ \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The force on a charge particle in a magnetic field is governed by the vector cross product, mathematically relying on the sine of the incident angle.
Formula:
$$ F = qvB \sin(\theta) $$
Solution:
To maximize the force \( F \), the trigonometric function \( \sin(\theta) \) must be at its maximum positive value.
The maximum value of the sine function is \( 1 \).
This occurs exactly when the angle \( \theta = 90^\circ \) (perpendicular entry).
Thus, entering the field perpendicularly results in maximum deflection force.
Why other options are incorrect:
Option C results in \( 0 \) force. Options A and D result in intermediate fractional forces (\( \frac{\sqrt{3}}{2} \) and \( \frac{\sqrt{2}}{2} \) of the maximum, respectively).
When a charged particle enters the magnetic field parallel, then it will:
A
Deflect toward north
B
Deflect toward south
C
Move straight
D
Move in circular path
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
If a charge enters a magnetic field parallel to the field lines, the cross product of its velocity and the magnetic field is zero, resulting in no force.
Formula:
$$ F = qvB \sin(\theta) $$
Solution:
"Parallel" implies the angle \( \theta \) between velocity \( \mathbf{v} \) and magnetic field \( \mathbf{B} \) is \( 0^\circ \) (or \( 180^\circ \)).
Substitute this into the formula: \( \sin(0^\circ) = 0 \).
The calculated magnetic force is therefore zero.
According to Newton's First Law, with no net force, the particle continues moving in a straight line at constant speed.
Why other options are incorrect:
Options A, B, and D describe deflections, which can only happen if a non-zero magnetic force acts on the particle (requiring a non-zero angle).
The SI unit for magnetic induction B is tesla (T). 1 Tesla is equal to:
A
\( \text{NA}^{-1}\text{m}^{-1} \)
B
\( \text{NmA}^{-1} \)
C
\( \text{N}^{-1}\text{mA} \)
D
\( \text{NmA} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The Tesla is a derived SI unit defined by the magnetic force formula on a current carrying wire.
Formula:
$$ F = ILB \implies B = \frac{F}{IL} $$
Solution:
Write out the SI units for the variables in the rearranged formula: Force \( F \) is in Newtons (N), Current \( I \) is in Amperes (A), and Length \( L \) is in meters (m).
Imaginary lines which show imaginary magnetic field
D
Actual lines which show imaginary magnetic field
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Magnetic field lines are a visual tool introduced by Michael Faraday to conceptualize and map the invisible magnetic field.
Solution:
The magnetic field itself is a real physical entity that stores energy and exerts real forces.
However, the "lines" drawn to represent it do not exist physically in space; they are an illustrative abstraction.
Therefore, they are imaginary lines used to accurately depict a real (actual) physical field.
Why other options are incorrect:
Option B is false because you cannot touch or extract a "line" from a vacuum. Option C is false because the magnetic field itself is a real phenomenon. Option D is an oxymoron.
A charged particle entered in a magnetic field anti parallel to the field, magnetic force on this particle is:
A
\( \text{BINA} \)
B
\( \text{BeV}\sin(0) \)
C
Zero
D
\( \text{Iqlv} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The magnetic force requires a perpendicular velocity component relative to the magnetic field. Anti-parallel means they are exactly aligned but pointing in opposite directions.
Formula:
$$ F = qvB \sin(\theta) $$
Solution:
Anti-parallel implies the angle between the velocity vector and magnetic field vector is \( \theta = 180^\circ \).
Calculate the sine of this angle: \( \sin(180^\circ) = 0 \).
Substitute this into the force equation: \( F = qvB(0) = 0 \).
The resulting force is exactly Zero.
Why other options are incorrect:
Option A is the torque on a coil. Option B uses an angle of \( 0^\circ \) which equals zero, but anti-parallel strictly means \( 180^\circ \). Option D is dimensional gibberish.
Option B is the unit for the magnetic field (Tesla) itself. Options C is an incorrect dimension. Option D is specifically the unit of magnetic field strength, not flux.
Electromagnetism defines that magnetic fields are strictly the relativistic result of electric charges in motion (currents).
Solution:
A stationary charge creates an electrostatic field in space, but no magnetic field.
When a charge moves, it forms an electric current.
According to Biot-Savart Law and Ampere's Law, a current (moving charge) induces a circulating magnetic field around its path.
Therefore, only moving charges produce magnetic fields.
Why other options are incorrect:
Option B is false because stationary charges lack the necessary velocity. Option C is false for the same reason. Option D is too restrictive; moving electrons, protons, and other charged particles also produce magnetic fields, not just ions.
Magnetic flux is zero when angle between magnetic field and vector area is:
A
\( 0^\circ \)
B
\( 90^\circ \)
C
\( 60^\circ \)
D
\( 30^\circ \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Magnetic flux measures how much magnetic field "pierces" through a given surface. The vector area is defined as normal (perpendicular) to the surface plane.
Formula:
$$ \Phi = BA \cos(\theta) $$
Solution:
We want the flux \( \Phi \) to be zero. This requires \( \cos(\theta) = 0 \).
The cosine of an angle is zero when the angle is \( 90^\circ \).
Physically, this means the magnetic field lines are perfectly parallel to the surface plane (sliding across it), meaning none of them pierce through it.
Why other options are incorrect:
Option A yields maximum flux. Options C and D yield fractional flux values since \( \cos(60^\circ) = 0.5 \) and \( \cos(30^\circ) = 0.866 \).
The path of a neutron moving perpendicular to magnetic field is:
A
Curve
B
Straight line
C
Circle
D
Ellipse
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A magnetic field exerts a deflecting force to curve the path of a particle ONLY if that particle carries an electric charge.
Formula:
$$ F = qvB \sin(\theta) $$
Solution:
A neutron is an uncharged particle (\( q = 0 \)).
Therefore, regardless of its angle of entry (even perpendicular, \( \sin(90^\circ) = 1 \)), the force evaluates to \( F = 0 \).
According to Newton's First Law, an object in motion stays in motion in a straight line unless acted upon by a net external force.
Why other options are incorrect:
Options A, C, and D describe trajectories that require a continuous perpendicular force (centripetal force) acting on the object, which is impossible for a neutral particle in a magnetic field.
A charged particle of mass 'm' and charge 'q' is projected in a magnetic field of induction B at the angle '\( \theta \)'. The radius of curvature of its curved path given by:
A
\( r = \frac{mv}{qB} \)
B
\( r = \frac{mv}{qB\sin\theta} \)
C
\( r = \frac{mv\cos\theta}{qB} \)
D
\( r = \frac{mv\sin\theta}{qB} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
When a charged particle enters a magnetic field at an arbitrary angle \( \theta \), its velocity vector is resolved into parallel and perpendicular components. Only the perpendicular component causes the circular motion.
Formula:
$$ \frac{m(v_\perp)^2}{r} = q v_\perp B $$
Solution:
The perpendicular component of velocity is \( v_\perp = v \sin(\theta) \).
Equate the magnetic Lorentz force to the centripetal force: \( q(v \sin\theta)B = \frac{m(v \sin\theta)^2}{r} \).
Cancel one factor of \( (v \sin\theta) \) from both sides: \( qB = \frac{m(v \sin\theta)}{r} \).
Rearrange to solve for the radius \( r \): \( r = \frac{mv \sin\theta}{qB} \).
Why other options are incorrect:
Option A assumes a strictly perpendicular entry (\( 90^\circ \)). Option B incorrectly divides by the sine term. Option C incorrectly uses the parallel velocity component (cosine) which contributes to the helical pitch, not the radius.
The magnetic field inside the current carrying wire varies:
A
Inversely with r
B
Inversely with r\(^2\)
C
Directly with r
D
Directly with r\(^2\)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
For a thick, uniform current-carrying cylindrical wire, the magnetic field behavior differs inside the wire compared to outside the wire. This is determined using Ampere's Law.
Formula:
$$ \oint B \cdot dl = \mu_0 I_{\text{enc}} $$
Solution:
Inside the wire (at a radius \( r < R \), where \( R \) is wire radius), the current enclosed by an Amperian loop depends on the area: \( I_{\text{enc}} = I \frac{r^2}{R^2} \).
Magnetic flux is maximum when angle between magnetic field and vector area is:
A
\( 0^\circ \)
B
\( 90^\circ \)
C
\( 180^\circ \)
D
\( 45^\circ \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Magnetic flux measures the total magnetic field passing directly through a given surface area. Maximum flux occurs when the field passes perpendicular to the surface (which means parallel to the area's normal vector).
Formula:
$$ \Phi = BA \cos(\theta) $$
Solution:
The variable \( \theta \) is the angle between the magnetic field vector and the vector area (normal vector).
The mathematical function \( \cos(\theta) \) dictates the magnitude. It hits its maximum positive value of 1 when \( \theta = 0^\circ \).
Therefore, flux is maximized at \( 0^\circ \).
Why other options are incorrect:
Option B yields exactly zero flux (lines scrape the surface). Option C gives the maximum negative value (depending on sign convention, but 0 is standard for max positive). Option D gives a partial flux value.
Choose a circular Amperian loop of radius \( r \) strictly inside the toroidal core. The length of this loop is \( 2\pi r \).
If there are \( N \) total turns of wire, the total enclosed current is \( NI \).
Applying Ampere's law gives: \( B(2\pi r) = \mu_0 NI \).
Rearranging for \( B \) gives: \( B = \frac{\mu_0 NI}{2\pi r} \).
Why other options are incorrect:
Option A is the formula for the interior of a straight, infinitely long solenoid using turn density \( n \), not total turns \( N \). Options C and D are dimensionally incorrect.
Magnetic flux (\(\Phi\)) quantifies the total magnetic field passing perpendicularly through a given surface area.
Formula:
$$\Phi = \vec{B} \cdot \vec{A} = BA\cos\theta$$
Solution:
Option A (\(\text{T}\cdot\text{m}^2\)): Looking at the defining equation \(\Phi = B \cdot A\), the SI unit of magnetic flux density (\(B\)) is Tesla (\(\text{T}\)) and the unit of Area (\(A\)) is square meters (\(\text{m}^2\)), yielding \(\text{T}\cdot\text{m}^2\).
Option B (Weber): In the SI system, \(1\text{ T}\cdot\text{m}^2\) is given the special name Weber (Wb): $$1\text{ Wb} = 1\text{ T}\cdot\text{m}^2 = 1\text{ N}\cdot\text{m}\cdot\text{A}^{-1} = 1\text{ J}\cdot\text{A}^{-1}$$
Therefore, both \(\text{T}\cdot\text{m}^2\) and \(\text{Weber}\) are identical, correct SI representations of magnetic flux.
Option C (Both A and B) is the comprehensive correct answer.
Why other options are incorrect:
D (Tesla): Tesla (\(\text{T}\)) is the unit for magnetic flux density (magnetic field strength \(B\)), not magnetic flux (\(\Phi\)).
The Earth possesses a relatively weak intrinsic magnetic field that protects the atmosphere from solar wind.
Formula:
$$ \text{Earth's field range: } 25 \ \mu\text{T} \text{ to } 65 \ \mu\text{T} $$
Solution:
Expressed in Tesla, the Earth's field is about \( 0.25 \times 10^{-4} \ \text{T} \) to \( 0.65 \times 10^{-4} \ \text{T} \).
The unit Gauss (G) is related to Tesla by \( 1 \ \text{T} = 10^4 \ \text{G} \).
Converting the range to Gauss gives: \( 0.25 \ \text{G} \) to \( 0.65 \ \text{G} \).
Option A (\( 0.6 \ \text{G} \)) falls perfectly within this standard physical range.
Why other options are incorrect:
Options B, C, and D are extremely massive magnetic fields (MRI machines operate around 1.5 to 3 Tesla). Life on Earth would be vastly different if the field was on the order of full Teslas.
The macroscopic magnetic force exerted on a straight current-carrying wire situated in a uniform magnetic field depends on multiple physical parameters.
Formula:
$$ F = ILB \sin(\theta) $$
Solution:
From the formula, \( I \) represents the electric Current.
\( L \) represents the Length of the wire inside the field.
\( B \) represents the external Magnetic field strength.
Since the force scales linearly with all three of these parameters, it depends on all of them.
Why other options are incorrect:
Choosing only A, B, or C is incomplete because the force relies on the combination of all these factors.
Magnetic flux is maximum when angle between magnetic field and vector area is
A
\( 0^\circ \)
B
\( 90^\circ \)
C
\( 180^\circ \)
D
\( 45^\circ \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Magnetic flux measures the number of magnetic field lines passing perpendicularly through a surface. It is mathematically the dot product of the magnetic field vector and the area vector (which is normal to the surface).
Formula:
$$ \Phi = \mathbf{B} \cdot \mathbf{A} = BA \cos(\theta) $$
Solution:
The flux is maximized when the mathematical function \( \cos(\theta) \) reaches its maximum positive value.
The maximum value of \( \cos(\theta) \) is 1, which occurs exactly at \( \theta = 0^\circ \).
At this angle, the magnetic field is perfectly parallel to the area vector (meaning the field is hitting the surface perfectly face-on).
Why other options are incorrect:
Option B (\( 90^\circ \)) yields zero flux. Option C (\( 180^\circ \)) gives the maximum negative flux. Option D (\( 45^\circ \)) yields an intermediate value.
Two long, parallel conductors which are free to move are arranged 1.0 cm apart. A steady current of 20 A flows in each of the conductor in the same direction. The conductors
A
Remain stationary
B
Move towards each other
C
Move away from each other
D
Move at right angles to each other
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When two parallel wires carry currents, they exert a magnetic force on each other. If the currents are in the same direction, they attract.
Formula:
$$ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} $$
Solution:
The magnetic field created by one wire at the location of the other can be found using the right-hand rule.
Applying Fleming's left-hand rule (or \( \mathbf{F} = I(\mathbf{L} \times \mathbf{B}) \)) shows that the force on each wire points towards the other wire.
Because they are free to move, the attractive force will cause them to move towards each other.
Why other options are incorrect:
Option A ignores the existence of the magnetic force. Option C would happen if the currents were in opposite directions. Option D is physically impossible for parallel uniform fields.
A neutron having mass equal to a proton (\( m_p = 1.6 \times 10^{-27} \ \text{kg} \)) is moving in a magnetic field of intensity \( 1.20 \times 10^{-3} \ \text{T} \) with a speed of \( 2.0 \times 10^7 \ \text{ms}^{-1} \) what is the Maximum force experienced by the neutron.
A
\( 3.84 \times 10^{-15} \ \text{N} \)
B
\( 0 \ \text{N} \)
C
\( 3.84 \times 10^{12} \ \text{N} \)
D
\( 38.4 \times 10^{-15} \ \text{N} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The magnetic force acting on a particle depends fundamentally on the electric charge of that particle. Neutral particles do not interact with magnetic fields in this manner.
Formula:
$$ F = qvB \sin(\theta) $$
Solution:
Identify the particle: A neutron.
Recall the charge of a neutron: \( q = 0 \ \text{C} \).
Substitute this into the Lorentz force equation: \( F = (0) \cdot vB \sin(\theta) \).
Therefore, regardless of velocity or magnetic field strength, the magnetic force is exactly 0 N.
Why other options are incorrect:
Options A, C, and D are calculated by falsely assuming the neutron has the charge of a proton (\( 1.6 \times 10^{-19} \ \text{C} \)) and running the numbers.
\( \text{e/m} \) of an electron is given by the relationship,
A
\( \text{e/m} = \frac{2V}{B^2 r^2} \)
B
\( \text{e/m} = \left( \frac{V}{Br} \right)^2 \)
C
\( \text{e/m} = \frac{Vr}{B} \)
D
\( \text{e/m} = \frac{VB}{r} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In J.J. Thomson's experiment (or in a standard mass spectrometer setup), an electron is accelerated through a potential difference \( V \) and then enters a perpendicular magnetic field \( B \), moving in a circle of radius \( r \).
If the value of magnetic flux is \( 10 \ \text{Wb} \), when magnetic lines of force containing magnetic field strength of \( 1 \ \text{Tesla} \) passing through unit area of \( 10 \ \text{m}^2 \) then the angle between magnetic field and unit area is:
A
\( 180^\circ \)
B
\( 360^\circ \)
C
\( 90^\circ \)
D
\( 45^\circ \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Magnetic flux depends on the angle \( \theta \) between the magnetic field vector and the normal vector to the area.
Formula:
$$ \Phi = BA \cos(\theta) $$
Solution:
Identify the givens: Magnitude of flux \( |\Phi| = 10 \ \text{Wb} \), \( B = 1 \ \text{T} \), \( A = 10 \ \text{m}^2 \).
Substitute into the formula: \( 10 = (1)(10) \cos(\theta) \).
Solve for the cosine term: \( \cos(\theta) = 1 \) (or \( -1 \) if we consider the absolute magnitude of the flux).
Angles that satisfy \( |\cos(\theta)| = 1 \) are \( 0^\circ \) and \( 180^\circ \).
Since \( 0^\circ \) is not in the options, the correct answer must be \( 180^\circ \), which gives a flux of \( -10 \ \text{Wb} \) (magnitude is \( 10 \)).
Why other options are incorrect:
Option B (\( 360^\circ \)) is a full rotation yielding the same as \( 0^\circ \), but conventionally \( 180^\circ \) is used for anti-parallel vectors. Option C yields zero flux. Option D yields \( 7.07 \ \text{Wb} \).
A charge is projected with velocity of \( 10 \ \text{m/s} \) in a magnetic field of \( 10 \ \text{T} \) at angle of \( 60^\circ \). If force of \( 2.78 \times 10^{-17} \ \text{N} \) is exerted on the charge then value of charge will be:
A
\( 1.60 \times 10^{-19} \ \text{C} \)
B
\( 2.70 \times 10^{-19} \ \text{C} \)
C
\( 4.80 \times 10^{-19} \ \text{C} \)
D
\( 3.20 \times 10^{-19} \ \text{C} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The magnetic force on a charge moving at an angle \( \theta \) to a magnetic field is given by the Lorentz force equation.
Divide the force by this value: \( q = \frac{2.78 \times 10^{-17}}{86.6} = 0.0321 \times 10^{-17} \ \text{C} \).
Convert to scientific notation: \( q \approx 3.20 \times 10^{-19} \ \text{C} \) (which is exactly the charge of an alpha particle or two protons).
Why other options are incorrect:
Option A is the charge of a single electron/proton. Options B and C are mathematically incorrect derivations that occur if you fail to use the sine function or multiply incorrectly.
Due to current in straight conductor the distance between magnetic field lines.
A
Increases away from conductor
B
Increases towards conductor
C
Decreases away from conductor
D
Decreases and then increases towards conductor
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The magnetic field strength created by a long straight current-carrying wire is inversely proportional to the radial distance from the wire. Visually, stronger fields are represented by lines closer together.
Formula:
$$ B = \frac{\mu_0 I}{2\pi r} $$
Solution:
Observe the formula: \( B \propto \frac{1}{r} \). As distance \( r \) increases, the magnetic field strength \( B \) decreases.
In field line diagrams, a weaker field is depicted by field lines that are spaced further apart.
Therefore, as you move away from the conductor, the distance between the consecutive magnetic field lines increases.
Why other options are incorrect:
Option B implies the field gets weaker near the wire, which is false. Option C implies the field gets stronger away from the wire. Option D implies a complex oscillatory field which doesn't apply to a straight DC wire.
A long straight current carrying conductor has current direction from bottom to top when held vertically. What will be the direction of magnetic field lines when observed from below the conductor?
A
Clockwise
B
Vertically upward
C
Anticlockwise
D
Vertically downward
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The direction of magnetic field lines around a straight current-carrying conductor is determined by the Right-Hand Grip Rule. Perspective completely changes whether it appears clockwise or anticlockwise.
Formula:
$$ \text{Right-Hand Grip Rule} $$
Solution:
Point your right thumb in the direction of the current (bottom to top, i.e., upwards).
Your fingers curl around the wire in an anticlockwise direction if you were looking from the top down.
However, the question asks for the perspective from below the conductor looking upwards.
When viewed from below, an anticlockwise curl spatially translates to a clockwise observation.
Why other options are incorrect:
Option C is the perspective from above. Options B and D are impossible because the magnetic field forms closed loops around the wire, not straight lines parallel to it.
Two long straight parallel wires held vertically, have equal but opposite currents as shown in the figure. Which of the following effect will be observed?
A
Magnetic field at 'X' is stronger than that at 'Y' and 'Z'
B
Magnetic field at 'X', and 'Z' are same
C
Magnetic field at 'X' is weaker than that at Z
D
Magnetic field at 'X' is weaker than that at out, stronger than that at 'Z'
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
When two parallel wires carry currents in opposite directions, the magnetic fields generated by each wire reinforce one another in the space between them and partially cancel out in the regions outside them.
Let 'X' be the region directly between the two wires, and 'Y' and 'Z' be the regions outside.
Using the Right-Hand Rule, the field lines from both wires point in the same direction in region 'X', adding together constructively.
In regions 'Y' and 'Z', the field vectors from the two wires point in opposite directions, leading to a destructive superposition.
Therefore, the net magnetic field at 'X' is strictly stronger than at 'Y' and 'Z'.
Why other options are incorrect:
Options B, C, and D violate the superposition principle by falsely assuming the fields cancel between the wires or claiming the external fields are stronger.
Two long parallel wires Wire 1 and Wire 2 repel each other as shown in the figure. What could be the reasons?
A
Both carry current in same direction
B
Both carry current in opposite direction
C
Wire 1 has current, but Wire 2 has no current
D
Wire 2 has current, wire 1 has no current
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Parallel conductors carrying electric current exert magnetic forces on one another. The direction of the force depends on the relative directions of the currents.
Formula:
$$ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} $$
Solution:
According to the right-hand grip rule, a current produces a concentric magnetic field.
If currents are in the same direction, the magnetic fields between the wires cancel out (weaker field), causing them to attract.
If currents are in opposite directions, the magnetic fields between the wires reinforce each other (stronger field), causing the wires to repel.
Since the question states they repel, they must carry currents in opposite directions.
Why other options are incorrect:
Option A leads to attraction. Options C and D result in zero magnetic force because both wires must carry current to produce a mutual magnetic force.
The SI unit of magnetic flux is weber which is equal to:
A
\( \text{NmA}^{-1} \)
B
\( \text{Nm}^{2}\text{A}^{-1} \)
C
\( \text{Nm} \)
D
\( \text{AmA}^{-2} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Magnetic flux is defined as the dot product of the magnetic field and the area vector. Its SI unit is the Weber (Wb).
Formula:
$$ \Phi = B \cdot A $$
Solution:
Start with the units: \( \text{Wb} = \text{Tesla} \times \text{m}^2 \).
Recall that \( B = \frac{F}{IL} \), meaning \( 1 \ \text{Tesla} = 1 \ \frac{\text{N}}{\text{A \cdot m}} \).
Substitute this into the flux unit: \( \text{Wb} = \left( \frac{\text{N}}{\text{A \cdot m}} \right) \times \text{m}^2 \).
Simplify the expression: \( \text{Wb} = \frac{\text{N \cdot m}}{\text{A}} = \text{N m A}^{-1} \).
Why other options are incorrect:
Option B represents an incorrect dimensional analysis where the meter squared is not fully canceled. Option C is the unit for torque or work (Joule). Option D is dimensionally nonsensical for flux.
Magnetic field strength (or magnetic flux density) has the SI unit Tesla (\( \text{T} \)). It can be expressed in various derived SI units and relates to the CGS unit (Gauss).
Formula:
$$ \text{B} = \frac{\Phi}{A} = \frac{F}{IL} $$
Solution:
Evaluate Option C: \( \text{Tesla} = \frac{\text{Weber}}{\text{m}^2} \). This is correct.
Evaluate Option A: Using \( B = \frac{F}{IL} \), \( 1 \ \text{T} = \frac{\text{N}}{\text{A \cdot m}} = \text{N m}^{-1} \text{A}^{-1} \). Since \( 1 \ \text{T} = 1 \ \text{wb m}^{-2} \), this is also correct.
Evaluate Option B: The conversion factor between SI (Tesla) and CGS (Gauss) is exactly \( 1 \ \text{T} = 10^4 \ \text{G} \). This is correct.
Since all individual statements are correct, the answer is "All of these".
Why other options are incorrect:
No single option (A, B, or C) is incorrect; therefore, selecting only one of them would be incomplete.
A charge of two micro coulombs (\( 2 \ \mu\text{C} \)) moves with velocity of two meter per second (\( 2 \ \text{m/sec} \)) in the direction of two Tesla magnetic field. The force that will act on it will be:
A
\( 2 \ \text{N} \)
B
Zero
C
\( 8 \ \text{N} \)
D
\( 4 \ \text{N} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The magnetic force acting on a moving charge depends on the cross product of its velocity and the magnetic field. If the charge moves parallel to the magnetic field, no magnetic force acts on it.
Formula:
$$ F = qvB \sin(\theta) $$
Solution:
Identify the given values: \( q = 2 \times 10^{-6} \ \text{C} \), \( v = 2 \ \text{m/s} \), \( B = 2 \ \text{T} \).
Determine the angle \( \theta \): The phrase "in the direction of" means the velocity vector is parallel to the magnetic field vector, so \( \theta = 0^\circ \).
Calculate the sine of the angle: \( \sin(0^\circ) = 0 \).
Substitute into the formula: \( F = (2 \times 10^{-6})(2)(2)(0) = 0 \ \text{N} \).
Why other options are incorrect:
Options A, C, and D are the results of multiplying the magnitudes while incorrectly assuming \( \theta = 90^\circ \) (which would yield \( 8 \times 10^{-6} \ \text{N} \)) or ignoring the micro prefix.
Recall the standard expression for magnetic force on a single moving charge \( q \) with velocity \( \mathbf{v} \) in field \( \mathbf{B} \).
Match the formulation to the options provided. Option C correctly represents this vector cross product.
Why other options are incorrect:
Option A is the formula for magnetic flux. Option B is the magnetic field of a solenoid. Option D is the magnetic force on a current-carrying conductor, not a single point charge.
A stationary electric charge produces only an electric field in the space around it. However, when the charge is in motion, it constitutes an electric current, which generates a magnetic field in addition to its intrinsic electric field.
Identify the state of the charge: It is in uniform motion.
Recall that any charge inherently has an electric field.
Apply the principle of electromagnetism: moving charges create magnetic fields.
Therefore, it produces both fields simultaneously.
Why other options are incorrect:
Option A is true only for a stationary charge. Option B ignores the fundamental electric field that always exists around a charge. Option D is entirely false as charges always produce fields.
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