A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is R. When energy of particle is doubled, then new radius will be :

Question

A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is R. When energy of particle is doubled, then new radius will be : (A)  (R/√2)  (B)  2R  (C)  (R/2)  (D)  √2R

Tardigrade Admin · Accepted Answer

Force acting on charged particle moving in a magnetic fields is given by:

F =

qvB This is equal to the centripetal force acting on the body.

F = \frac{m v ^{2}}{R}

Hence

R = \frac{m v}{qB}

Given initial Radius

R = \frac{m v}{qB}

Now energy is doubled.
i.e.

\frac{1}{2} m v_{2}^{2} = 2 * \frac{1}{2} m v^{2}

i.e.

v_{2} = 2 v

Hence Radius

R_{2} = \frac{2 m v}{qB} = 2 R

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Q. A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is $R$ . When energy of particle is doubled, then new radius will be :

$\frac{R}{2}$

$2 R$

$\frac{R}{2}$

$2 R$

Q. A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is R. When energy of particle is doubled, then new radius will be :

2​R​

2R

2R​

2​R

Q. A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is $R$ . When energy of particle is doubled, then new radius will be :

$\frac{R}{2}$

$2 R$

$\frac{R}{2}$

$2 R$