Question

derive an expression for radius of circle cover by charged particle upon entering uniform magnetic field perpendicular​

Answer 1

To derive:

Expression for radius of circle covered by a charged particle after entering a uniform magnetic field perpendicular to the velocity of the charge.

Derivation:

Let us consider that a charged particle is moving with velocity v and enters a magnetic field of magnitude B.

The magnetic force will provide the centripetal component while undergoing the circular trajectory.

$\sf{ \therefore \: F_{M} = F_{C}}$

$\sf{ = > \: q \bigg \{v \times B \times \sin(90 \degree) \bigg \} = \dfrac{m {v}^{2} }{r} }$

$\sf{ = > \: q \bigg \{v \times B \times 1\bigg \} = \dfrac{m {v}^{2} }{r} }$

$\sf{ = > \: q \times v \times B = \dfrac{m {v}^{2} }{r} }$

$\sf{ = > \: r = \dfrac{mv }{qB} }$

So, final expression:

$\boxed{ \red{ \large{\rm{ \: r = \dfrac{mv }{qB} }}}}$