Question

An electron with charge -e and mass m travels at a speed v in a plane

perpendicular to a magnetic field of magnitude B. The electron follows a

circular path of radius R. In a time, t, the electron travels halfway around the

circle. What is the amount of work done by the magnetic field?​

Answer 1

Couldn't understand the question

Answer 2

Given:

An electron with charge -e and mass m travels at a speed v in a plane perpendicular to a magnetic field of magnitude B. The electron follows a circular path of radius R. In a time, t, the electron travels halfway around the circle.

To find:

Work done by magnetic field ?

Calculation:

In the circular trajectory of the electron, the magnetic force is responsible for providing the centripetal component of acceleration to the moving electron.

So, the force vector is is directed radially inwards and is at 90° with the displacement vector .

So, let net work be W :

$\therefore \: W = \vec{F} \: . \: \vec{d}$

$\implies \: W = | \vec{F}| \times | \vec{d}| \times \cos( \theta)$

$\implies \: W = | \vec{F}| \times | \vec{d}| \times \cos( {90}^{ \circ} )$

$\implies \: W = 0 \: joule$

• Irrespective of the time travelled in the circular path, the net work will always be zero.

So, net work done is zero joule.