Question

A proton travels with a speed of 1.8×106 m/s at an angle 53◦ with a magnetic field of 0.49 T pointed in the +y direction. The mass of the proton is 1.672 × 10−27 kg. What is the magnitude of the magnetic force on the proton?

Answer 1

Magnetic force,
`F=1.12* 10^(-13)\ N`

Step-by-step explanation:

It is given that,

Velocity of proton,
`v=1.8* 10^6\ m/s`

Angle between velocity and the magnetic field, θ = 53°

Magnetic field, B = 0.49 T

The mass of proton,
`m=1.672* 10^(-27)\ kg`

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

The magnitude of magnetic force is given by :

`F=qvB\ sin\theta`

`F=1.6* 10^(-19)\ C* 1.8* 10^6\ m/s* 0.49\ Tsin(53)`

`F=1.12* 10^(-13)\ N`

So, the magnitude of the magnetic force on the proton is
`1.12* 10^(-13)\ N`. Hence, this is the required solution.