Question

An electron moving with a speed of 6.2×105 m/sm/s in the positive x direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 3.4×10−13 N that points in the positive z direction.What is the magnitude of the magnetic field?

Answer 1

Answer:

The magnitude of the magnetic field is 3.42 T.

Explanation:

Given that,

Speed of the electron, $v=6.2\times 10^5\ m/s$

The force acting on the particle in the magnetic field, $F=3.4\times 10^{-13}\ N$

To find,

The magnitude of the magnetic field.

Solution,

There exists a magnetic force when an electron is moving in a magnetic field. The magnetic force is given by :

$F=qvB\ sin\theta$

Here, q = -e

$F=evB$

$B=\dfrac{F}{ev}$

$B=\dfrac{3.4\times 10^{-13}}{1.6\times 10^{-19}\times 6.2\times 10^5}$

B = 3.42 T

So, the magnitude of the magnetic field is 3.42 T.