An $\alpha$-particle with a specific charge of $2.5 imes 10^{7}\ C\ {kg}^{- 1}$ moves with a speed of $2 imes 10^{5}\ {ms}^{- 1}$ in a perpendicular magnetic field of 0.05 T. Then the radius of the circular path described by it is

Question

An $\alpha$-particle with a specific charge of $2.5 	imes 10^{7}\ C\ {kg}^{- 1}$ moves with a speed of $2 	imes 10^{5}\ {ms}^{- 1}$ in a perpendicular magnetic field of 0.05 T. Then the radius of the circular path described by it is

Dr. Abubacker · Accepted Answer

(c) $\ \ \ \ \ \ \ \ r = \frac{mv}{Bq}$
$\Rightarrow \ \ r = \frac{v}{B\frac{q}{m}} = \frac{2 	imes 10^{5}}{0.05 	imes 2.5 	imes 10^{7}}$
$\ \ \ \ \ \ \ \ \ \  = \frac{2 	imes 10^{7}}{12.5 	imes 10^{7}} = \frac{200}{12.5}\ cm = 16\ cm$

Q. An $\alpha$-particle with a specific charge of $2.5 \times 10^{7}\ C\ {kg}^{- 1}$ moves with a speed of $2 \times 10^{5}\ {ms}^{- 1}$ in a perpendicular magnetic field of 0.05 T. Then the radius of the circular path described by it is

Explanation

Reviewed by Dr. Abubacker

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