Question

A particle of charge $q$ and mass $m$ is moving at a speed $v$ enters a uniform magnetic field of strength $B$ as shown below.
How much work is done by the magnetic field on the charge as the field accelerates the charge into a circle of radius $r$?

• A. $0$
• B. $\frac{1}{2}m{v}^2$
• C. $qvBr$
• D. $m{v}^2$
• E. Cannot be determined

Answer 1

The correct option is A $0$

As the charge is moving perpendicular to the magnetic field, therefore, direction of force on charge by magnetic field will be given by Fleming's left hand rule, that is vertically downward (perpendicular to both velocity and magnetic field).

now work done is given by,

$W=Fd\cos \theta$ ,

but $\theta ={90}^0$, as force and displacement (velocity) are perpendicular to each other,

therefore, $W=Fd\cos {90}^0=0$