First slide

Force on a charged particle moving in a magnetic field

Question

A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51 T as shown in figure. The angle $θ$ through which the proton deviates from the initial direction of its motion, is

Moderate

A

15⁰
B

30⁰
C

45⁰
D

60⁰

Solution

According to following figure, $\sin θ = \frac{d}{r}$
$\begin{array}{l} Also, r = \frac{\sqrt{2 mK}}{qB} = \frac{1}{B} \sqrt{\frac{2 mV}{q}} \\ ∴ \sin θ = Bd \sqrt{\frac{q}{2 mV}} \\ = 0 .51 \times 0 .1 \sqrt{\frac{1 .6 \times 10^{- 19}}{2 \times 1 .67 \times 10^{- 27} \times 500 \times 10^{3}}} \\ = \frac{1}{2} \Rightarrow θ = 30^{0} \end{array}$

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