Question

An electron moves with a speed of 5.0 x 10^4m/s

Answer 1

Answer:

$F=1.6\times 10^{-15}\ N$

Explanation:

Given that,

Speed of the electron, $v=5\times 10^4\ m/s$

Magnetic field, B = 0.20 T

We need to find the magnitude of magnetic force on the electron. The formula for the magnetic force is given by :

$F=qvB\ sin\theta$

Here, $\theta=90^{\circ}$

$F=qvB$, q is the charge on electron

$F=1.6\times 10^{-19}\times 5\times 10^4\times 0.2$

$F=1.6\times 10^{-15}\ N$

So, the magnitude of the magnetic force on the electron is $1.6\times 10^{-15}\ N$. Hence, this is the required solution.