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

`B = 1.03 * 10^(-4) T`

Step-by-step explanation:

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

`F = q(\vec v * \vec B)`

so we will have

`F = qvBsin\theta`

now we will have

`F = 4.6 * 10^(-3) N`

`q = 8.1 \mu C`

`\theta = 50^o`

`v = 7.2 * 10^6 m/s`

now we have

`4.6 * 10^(-3) = (8.1 * 10^(-6))(7.2 * 10^6)Bsin50`

`B = 1.03 * 10^(-4) T`