First slide

Force on a charged particle moving in a magnetic field

Question

A charged particle is projected with velocity v in a uniform magnetic field making an angle of $45^{0}$ with the field. The particle followed a helical path of radius 10 cm. If the same particle is projected with velocity 2V in the same magnetic field making an angle of $60^{0}$ with the field, radius of helical path followed by the particle will be

Moderate

A

$10 \sqrt{6} c m$
B

$10 \sqrt{3} c m$
C

$20 \sqrt{6} c m$
D

10 cm

Solution

(1)
$\begin{array}{l} r = \frac{V \sin θ}{ω} = \frac{V \sin θ}{(Q B / m)} \\ ∴ r \propto V \sin θ \\ ∴ \frac{r}{10} = \frac{(2 V) \sin 60^{0}}{V \sin 45^{0}} \Rightarrow r = 10 \sqrt{6} c m \end{array}$

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