Question

A proton, traveling with a velocity of 4.5 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.0 × 10−14 N and a direction of due south. What are the magnitude and direction of the magnetic field causing the force?

Answer 1

Magnetic field, B = 0.11 i (in k direction)

Step-by-step explanation:

It is given that,

Velocity of the proton,
`v=4.5* 10^6\ m/s`

Magnetic force,
`F=8* 10^(-14)\ N` (due south)

The magnetic force acting on the electron is given by :

`F=qvB`

(-j) = (i) × (B)

`B=(F)/(qv)`

q is the charge on proton

`B=(8* 10^(-14)\ N)/(1.6* 10^(-19)\ C* 4.5* 10^6\ m/s)`

B = 0.11

So, the magnitude of 0.11 T is acting on the proton in and is acting directed upward above the plane. Hence, this is the required solution.