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.