Question

An electron moving horizontally to the right with speed v enters a region where a uniform magnetic field exists pointing into the page with magnitude B. As a result the electron executes uniform circular motion with radius r. What is the total work done by the magnetic field during half cycle of the electron's motion in the field? A Br/ e B -Br / e C. 2pireBv D 2pir / eBv E 0

Answer 1

Answer:E

Step-by-step explanation:

Given

speed of electron is v

Magnitude of magnetic field B

radius of circular motion

Work done by Magnetic field is given by

`W=q\left ( \vec{v}* \vec{B}\right )\cdot d`

where
`\vec{v}* \vec{B}` will yield a vector perpendicular to both magnetic field and velocity vector .

`\vec{v}* \vec{B}` will also perpendicular to displacement d so net work done will be zero

`W=qvBd\cos 90`

`W=0`