Question

A charge of -8.1 µC is traveling at a speed of 7.2 106 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 50°. A force of magnitude 4.6 10-3 N acts on the charge. What is the magnitude of the magnetic field?

Answer 1

Answer:

$B = 1.03 \times 10^{-4} T$

Explanation:

As we know that magnetic force on a moving charge is given by the formula

$F = q(\vec v \times \vec B)$

so we will have

$F = qvBsin\theta$

now we will have

$F = 4.6 \times 10^{-3} N$

$q = 8.1 \mu C$

$\theta = 50^o$

$v = 7.2 \times 10^6 m/s$

now we have

$4.6 \times 10^{-3} = (8.1 \times 10^{-6})(7.2 \times 10^6)Bsin50$

$B = 1.03 \times 10^{-4} T$