Answer to Question #196036 in Electric Circuits for Kenma

Question

An electron moving at 8.21 × 10³ m/s in a 6.76-T magnetic field experiences a magnetic force of 4.75×10⁻¹⁶ N. What angle does the velocity of the electron make with the magnetic field?

answer to 2 decimal places

Answer 1

The magnitude of force is calculated from the relation:

$$F=qvBsin\theta$$

where

$q= charge\\F=force\\v=velocity\\B=magnetic\ field\ strength\\\theta=angle\ between\ direction\ of \ velocity\ and\ magnetic\ field\ strength$

So,

$F=qvBsin\theta\\\Rightarrow sin\theta=\dfrac{F}{qvB}=\dfrac{4.75\times10^{-16}}{(1.6\times10^{-19})(8.21\times10^3)(6.76)}=0.0535$

So,

$\theta=sin^{-1}(0.535)=32.34\degree$

Hence, the velocity of the electron makes an angle of $32.34\degree$with the magnetic field.