Question

Electrons and protons travel from the Sun to the Earth at a typical velocity of 4.06 ✕ 105 m/s in the positive x-direction. Thousands of miles from Earth, they interact with Earth's magnetic field of magnitude 2.99 ✕ 10−8 T in the positive z-direction. Find the magnitude and direction of the magnetic force on a proton. Find the magnitude and direction of the magnetic force on an electron.

Answer 1

Answer:

(a). The magnitude of the magnetic force on a proton is $19.423\times10^{-22}\ N$

The direction of force in negative y direction.

(b). The magnitude of the magnetic force on a proton is $19.423\times10^{-22}\ N$

The direction of force in positive y direction.

Explanation:

Given that,

Velocity $v=4.06\times10^{5}\ m/s$

Magnetic field $B=2.99\times10^{-8}\ T$

(a). For proton,

We need to calculate the force

Using formula of force

$F=q(v\times B)$

Put the value into the formula

$F=1.6\times10^{-19}\times4.06\times10^{5}\hat{i}\times2.99\times10^{-8}\hat{k}$

$F=-19.423\times10^{-22}\hat{j}\ N$

Negative sign shows the direction of force in negative y direction

The magnitude of the magnetic force on a proton is $19.423\times10^{-22}\ N$

(b). For electron,

Using formula of force

$F=q(v\times B)$

Put the value into the formula

$F=-1.6\times10^{-19}\times4.06\times10^{5}\hat{i}\times2.99\times10^{-8}\hat{k}$

$F=19.423\times10^{-22}\hat{j}\ N$

The direction of force in positive y direction

The magnitude of the magnetic force on a proton is $19.423\times10^{-22}\ N$

Hence, This is the required solution.