First slide

Force on a charged particle moving in a magnetic field

Question

AIfa particles (m= 6.7x10^-27 kg and q = 2e) arc accelerated from rest through a potential difference of 6 . 7 kV. Then they enter a magnetic field B = 0 .2 T perpendicular to their direction of motion. The radius of the path described by them is

Moderate

A

8.375m
B

8.375 cm
C

$\infty$
D

none of the above

Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = qV or v = {[\frac{2 qV}{m}]}^{1 / 2} \\ r = \frac{mV}{qB} = \frac{m}{qB} \sqrt{(\frac{2 qV}{m})} = \frac{1}{B} \sqrt{(\frac{2 Vm}{q})} \\ = \frac{1}{0 \cdot 2} [\frac{{2 \times (6 \cdot 7 \times 10^{3}) \times 6 \cdot 7 \times 10^{- 27}]}^{1 / 2}}{2 \times 1 \cdot 6 \times 10^{- 19}}] \\ = 8 \cdot 375 cm \end{array}$

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