Question

A charge of −8.3 μC is traveling at a speed of 7.4 × 106 m/s in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is 52°. A force of magnitude 5.4 × 10−3 N acts on the charge. What is the magnitude of the magnetic field?

Answer 1

Answer:

The magnitude of the magnetic field is   $1.1\times 10^{-4}\, T$

Explanation:

Given charge q = −8.3 μC

speed of charge $v=7.4\times 10^{6}\, \frac{m}{s}$

Angle between magnetic field and the velocity of charge $\Theta =52^{\circ}$

Strength of magnetic force on charge $F=5.4\times 10^{6}\, N$

Let magnitude of magnetic field be B

Since the magnetic force on a free moving charge in the magnetic field given by

$F=\left |q\vec{v}\times \vec{B}\right |=\left | qvBsin\Theta\right |$

=>$5.4\times 10^{-3}=\left | -8.3\times 10^{-6}\times 7.4\times 10^{6}\times sin(52^{\circ})\times B \right |$

=>$B=1.1\times 10^{-4}\, T$

Thus the magnitude of the magnetic field is   $1.1\times 10^{-4}\, T$