Question

Force on a proton of charge 2e in a magnetic field of B at 45 degree while moving with 2 m/s is

Answer 1

A proton with a charge 2e is moving in a magnetic field of intensity B at an angle of 45° with velocity 2 m/s.

The force experienced by the proton is :

$\vec{F} = q( \vec{v} \times \vec{B})$

$\implies | \vec{F}| = q \times v \times B \times \sin( \theta)$

$\implies | \vec{F}| = 2e \times 2 \times B \times \sin( {45}^{ \circ} )$

$\implies | \vec{F}| = 4eB \times \dfrac{1}{ \sqrt{2} }$

$\implies | \vec{F}| = 2 \sqrt{2} eB$

So, force experienced is 2√2 (eB).