Question

A proton enters a uniform magnetic field of 5T with velocity 4×107 ms-1 at right angles to the field. The magnetic force acting on the proton is

Answer 1

Answer: The magnetic force on the proton is $F = 3.204\times10^{-11}\ N$

Explanation:

Given that,

Charge on the proton $q= 1.602\times10^{-19}C$

Magnetic field B = 5 T

Velocity $v = 4\times10^7 m/s$

The magnetic force is

$F = q lvB\ sin\ \theta$

Now, put the values in equation (I)

$F = 1.602\times10^{-19}\times 4\times10^7\times 5 \times sin\ 90^{0}$

$F = 3.204\times10^{-11}\ N$

Hence, The magnetic force on the proton is $F = 3.204\times10^{-11}\ N$