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## Q. 3.9

A charged particle having charge thrice that of an electron and mass twice that of an electron is accelerated through a potential difference of 50 V before it enters a uniform magnetic field of flux density of $0.02 Wb/m^{2}$ at an angle of 25° with the field. Find (i) the velocity of the charged particle before entering the field, (ii) radius of the helical path, and (iii) time of one revolution.

## Verified Solution

The charge of the particle is, Q = 3 q and the mass of the particle is, M = 2 m. (i) The velocity of the charged particle before entering the field is,

$v=\sqrt{\frac{2aV}{M} } \sqrt{\frac{2(3q)V}{2m} }=\sqrt{\frac{6qV}{2m} }$

$=\sqrt{\frac{6\times (1.602\times 10^{-19} )\times 50}{2\times 9.1\times 10^{-31} } }$

$= 5.14 × 10^{6} m/s.$

(ii) The radius of the helical path is,

$r=\frac{M v \sin \theta }{Q B}=\frac{2mv\sin \theta }{3qB}$

$=\frac{2\times 9.1\times 10^{-31}\times 5.14\times 10^{6}\times \sin 25^{\circ } }{0.02\times 3\times 1.602\times 10^{-19} }$

= 0.411 mm.

(iii) Time for one revolution,

$T=\frac{2\pi M}{BQ}=\frac{2\pi (2m)}{B(3q)}$

$=\frac{4\pi \times 9.1\times 10^{-31} }{0.02\times 3\times 1.602\times 10^{-19} }$

$= 1.19 × 10^{–9} s$