Question

A proton moves perpendicular to a uniform magnetic field B S at a speed of 1.00 3 107 m/s and experiences an acceleration of 2.00 3 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.

Answer 1

Step-by-step explanation:

It is given that,

Speed of proton,
`v=10^7\ m/s`

Acceleration of the proton,
`a=2* 10^(13)\ \ m/s^2`

The force acting on the proton is balanced by the magnetic force. So,

`ma=qvB\ sin(90)`

`B=(ma)/(qv)`

m is the mass of proton

`B=(1.67* 10^(-27)\ kg* 2* 10^(13)\ \ m/s^2)/(1.6* 10^(-19)* 10^7\ m/s)`

B = 0.020875

or

B = 0.021 T

So, the magnitude of magnetic field is 0.021 T. As the acceleration in +x direction, velocity in +z direction. So, using right hand rule, the magnitude of B must be in -y direction.