Question

A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength of the magnetic field if there is a 45° angle between it and the proton's velocity?

Answer 1

Answer:

Magnetic field, $B=4.16\times 10^{-5}\ T$

Explanation:

It is given that,

Velocity of proton, $v=3.5\times 10^7\ m/s$

Magnetic force, $F=1.65\times 10^{-16}\ N$

Charge of proton, $q=1.6\times 10^{-19}\ C$

We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :

$F=qvB\ sin\theta$

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}$

B = 0.0000416 T

$B=4.16\times 10^{-5}\ T$

Hence, this is the required solution.