Question

A charge 3.2 x 10^-19 C moving with velocity 10^6 m/s
enters in a magnetic field of intensity 3 Wb/m ^2at an
angle of 60°. Calculate the force acting on the charge.
Write the necessary law for finding the direction of
magnetic force acting on a moving charge particle in a
magnetic field.
​

Answer 1

Answer:

Charge = 3.2 × 10^-19 C

velocity = 10^6 m/s

Mag field intensity = 3 wb / ^2

at angle 60°

Force = q ( V vector × B vector )

F = qvBsinθ

( sin60° =

$\frac{ \sqrt{3} }{2}$

$force \: = 3.2 \times {10}^{ - 19} \times {10}^{6} \times 3 \times \frac{ \sqrt{3} }{2}$

now calculate it

$4.8 \sqrt{3} \times 10 {}^{ - 13}$

direction of magnetic field can be determined by right hand thumb rule.