Physics
46 Solved Past Papers
2011 – 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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The SI unit of magnetic flux is weber. Weber can also be expressed as ____.
[SZABMU 2024]
A
Joule per ampere
B
Joule per coulomb
C
Newton per ampere
D
Newton per coulomb
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Units in electromagnetism can be heavily intertwined using work, force, and current relationships. The Weber can be derived down to energy and current.
Formula:
$$ \Phi = B \cdot A \quad \text{and} \quad W = F \cdot d $$
Solution:
Start with the base definition: \( 1 \ \text{Weber} = 1 \ \text{Tesla} \cdot \text{m}^2 \).
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.
Which of the following statement is incorrect for any magnetic field lines?
[UHS 2024]
A
Lines start at north pole and ends at south pole
B
The lines are curved
C
Lines never touch or cross each other
D
Magnetic field is strongest when lines are farthest
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Magnetic field lines are visual aids whose spacing visually indicates the relative magnitude of the magnetic field vector in that region of space.
Solution:
By convention, the density of magnetic field lines (how closely packed they are) directly correlates to the strength of the magnetic field.
Where lines are very close together, the field is remarkably strong (like near the poles of a magnet).
Conversely, where lines are farthest apart, the magnetic field is at its weakest.
Therefore, the statement "Magnetic field is strongest when lines are farthest" is fundamentally incorrect.
Why other options are incorrect:
Options A, B, and C are all true, factual properties of standard magnetic field mapping, meaning they are not the "incorrect" statement being asked for.
The formula \( \Phi = \mathbf{B} \cdot \mathbf{A} \) represents
[UHS 2024]
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:
[NUMS 2024]
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.
Magnetic field lines set up in the surrounding of current carrying wire will be:
[BUMHS 2024]
A
Circular
B
Along the current
C
Radially outward
D
Opposite to current
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Hans Christian Ørsted discovered that a straight current-carrying conductor generates a magnetic field in the surrounding three-dimensional space, the shape of which is determined by the Right-Hand Grip Rule.
Solution:
If you grasp a straight wire with your right hand such that your thumb points in the direction of the conventional current...
Your fingers will naturally curl around the wire.
This curl physically traces the path of the magnetic field lines, forming closed, concentric circular loops centered on the wire axis.
Why other options are incorrect:
Options B and D imply the field lines are straight and parallel to the wire, which is entirely false. Option C describes the topology of an electric field extending from a line of static charge, not a magnetic field.
Electrons of mass m and charge e are accelerated through a potential difference V and strike the target. The maximum speed of these electrons is:
[BUMHS 2024]
A
\( \sqrt{\frac{2eV}{m}} \)
B
\( \frac{eV}{m} \)
C
\( \sqrt{\frac{eV}{m}} \)
D
\( \frac{eV^2}{m} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
When a charged particle is accelerated across a voltage gap (potential difference), the electrical work done on the particle is completely converted into its kinetic energy (assuming starting from rest).
Formula:
$$ W = \Delta K.E \implies qV = \frac{1}{2}mv^2 $$
Solution:
Substitute the specific charge of an electron: \( q = e \). The electrical energy given to the electron is \( E = eV \).
Equate this to the classic formula for kinetic energy: \( eV = \frac{1}{2}mv^2 \).
Multiply both sides by 2 to clear the fraction: \( 2eV = mv^2 \).
Divide by mass \( m \): \( v^2 = \frac{2eV}{m} \).
Take the square root to isolate the maximum velocity \( v \): \( v = \sqrt{\frac{2eV}{m}} \).
Why other options are incorrect:
Option C forgets the factor of 2 originating from the \( \frac{1}{2} \) in the kinetic energy equation. Options B and D lack the necessary square root required to solve for \( v \) from \( v^2 \).
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:
[UHS 2023]
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?
[UHS 2023]
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:
[SZABMU 2023]
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?
[SINDH 2023]
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?
[SINDH 2023]
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:
[NUMS 2023]
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:
[NUMS 2023]
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 (\( \text{kg s}^{-2} \text{A}^{-1} \)).
In Fleming's right-hand rule, the second finger indicates:
[NUMS 2023]
A
Force
B
Magnetic field
C
Induced current
D
Motion
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Fleming's Right-Hand Rule is a mnemonic used to determine the direction of induced current when a conductor moves through a magnetic field (generator effect).
Solution:
The Thumb represents the direction of the applied Motion (or Force).
The First (Index) finger points in the direction of the Magnetic Field (North to South).
The Second (Middle) finger is extended orthogonally to represent the direction of the Induced current.
Why other options are incorrect:
Option A and D refer to the thumb. Option B refers to the first (index) finger.
Electric forces change the magnitude and direction of velocity while magnetic forces change _ of velocity.
[UHS 2022]
A
Only magnitude
B
Only direction
C
Magnitude and direction
D
Neither magnitude nor direction
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Electric forces can do work on a charge, speeding it up or slowing it down. Magnetic forces, however, always act perpendicular to the velocity vector of the charge.
Because the magnetic force is always perpendicular to velocity, the dot product of force and velocity is zero.
This means magnetic forces do absolutely no mechanical work (\( W = 0 \)) on a moving charge.
By the work-energy theorem, if no work is done, kinetic energy (and thus speed or velocity magnitude) remains constant.
The perpendicular force acts strictly as a centripetal force, altering only the direction of the velocity vector.
Why other options are incorrect:
Option A is false because magnetic fields cannot change speed. Option C conflates magnetic forces with electric forces. Option D is false because a force must cause an acceleration, which is a change in velocity (direction in this case).
Which surface has greater magnetic flux in same magnetic field, each has an area \( 1\text{m}^2 \)?
[UHS 2022]
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:
[SZABMU 2022]
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:
[SZABMU 2022]
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:
[SZABMU 2022]
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:
[ETEA 2022]
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.
The path of a neutron moving perpendicular to magnetic field is:
[DUHS 2022]
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:
[DUHS 2022]
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:
[NUMS 2022]
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:
[NUMS 2022]
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
[NMDCAT 2020]
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
[MDCAT 2018]
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.
[MDCAT 2018]
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,
[MDCAT 2018]
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:
[MDCAT 2017]
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:
[MDCAT 2017]
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.
The SI unit of magnetic flux is weber which is equal to:
[ETEA 2011]
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.
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