Question

A proton has a charge of 1.6 x 10-19 C and moves perpendicular to a uniform magnetic field of 2 T at a speed of 2.5 x 106 m/s. What is the force experienced by the proton? What will be its acceleration?

Answer 1

Step-by-step explanation:

It is given that,

Charge on proton,
`q=1.6* 10^(-19)\ C`

Magnetic field, B = 2 T

Speed of proton,
`v=2.5* 10^6\ m/s`

To find,

The force experienced by proton and its acceleration.

Solution,

The force experienced by a proton is given by :

`F=qvB`

`F=1.6* 10^(-19)* 2.5* 10^6* 2`

`F=8* 10^(-13)\ N`

We know that the product of mass and acceleration is equal to force. If a is the acceleration of proton. So,

`a=(F)/(m)`

`a=(8* 10^(-13))/(1.67* 10^(-27))`

`a=(4.79* 10^(14))\ m/s^2`