Question

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, with a magnitude of 0.08 T. Find the force (magnitude and direction) on the charge.

Answer 1

Step-by-step explanation:

It is given that,

Magnitude of charge,
`q=15\ \mu C=15* 10^(-6)\ C`

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field,
`B=0.08\ j`

Velocity,
`v=(5\ cos25)i+(5\ sin25)j`

`v=[(4.53)i+(2.11)j]\ m/s`

We need to find the magnitude of force on the charge. Magnetic force is given by :

`F=q(v* B)`

`F=15* 10^(-6)[(4.53i+2.11j)* 0.08\ j]`

Since,
`i* j=k\ and\ j* j=0`

`F=15* 10^(-6)[(4.53i)* (0.08)\ j]`

`F=0.00000543\ kN`

`F=5.43* 10^(-6)\ kN`

So, the force acting on the charge is
`5.43* 10^(-6)\ kN` and is moving in positive z axis. Hence, this is the required solution.