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?

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Answer 1

Answer:

The magnetic force on the electrom provides the centripetal force, so, we have,

Bqv=

R

mv

2

⇒R=

Bq

mv

So, if velocity is doubled and magnetic field intensity is halved, then the radius will become 4 times.

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

.

Explanation:

The magnetic force acts in such a way that the direction of the magnetic force and velocity are always perpendicular to each other. If force and velocity are perpendicular force and displacement are also perpendicular, thus W= FS cos q, if q = 90, work done will be zero.