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

Explanation:

Given

speed of electron is v

Magnitude of magnetic field B

radius of circular motion

Work done by Magnetic field is given by

$W=q\left ( \vec{v}\times \vec{B}\right )\cdot d$

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

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

$W=qvBd\cos 90$

$W=0$