Question

A particle of charge $1\mu C$ is at rest in a magnetic field $\overset{\to }{\mathop{B}}\,=-2\hat{k}\text{ }tesla$. Magnetic Lorentz force on the charge particle with respect to an observer moving with velocity $\overset{\to }{\mathop{v}}\,=-5\hat{i}m{{s}^{-1}}$ will be

A) Zero

B)$-{{10}^{5}}\hat{j}N$

C) $-{{10}^{6}}\hat{j}N$

D) ${{10}^{5}}\hat{j}N$

Answer 1

We know that the Lorentz force on a charged particle is given by: ${{F}_{l}}=q\left( v\times B \right)$. Apply the formula for $q=1\mu C$, $\overset{\to }{\mathop{B}}\,=-2\hat{k}\text{ }tesla$ and $\overset{\to }{\mathop{v}}\,=+5\hat{i}m{{s}^{-1}}$ to find the Lorentz force on the charged particle.

Complete step by step answer:

We have the following data as:

$q=1\mu C$

$\overset{\to }{\mathop{B}}\,=-2\hat{k}\text{ }tesla$

Velocity of observer $=-5\hat{i}m{{s}^{-1}}$

Velocity of charged particle (${\mathop{v}}$)= $0-(-5\hat{i}m{{s}^{-1}})$ =$(+5\hat{i}m{{s}^{-1}})$

So, by applying the formula for Lorentz Force: ${{F}_{l}}=q\left( v\times B \right)$

We get:

${{F}_{l}}=1\mu C\left( +5\hat{i}m{{s}^{-1}}\times -2\hat{k}\text{ }tesla \right)$

$={{10}^{-6}}\times -10\left( \hat{i}\times \hat{k} \right)$

$={-{10}^{-5}}\left( \hat{i}\times \hat{k} \right)......(1)$

As we know that: $\left( \hat{i}\times \hat{k} \right)=-\hat{j}$

So, we get:

${{F}_{l}}={{10}^{-5}}\hat{j}N$

So, the correct answer is “Option D”.

Note:

In physics (specifically in electromagnetism) the Lorentz force (or electromagnetic force) is the combination of the electric and magnetic force on a point charge due to electromagnetic fields. A particle of charge q moving with a velocity v in an electric field E and a magnetic field B experiences a force.

The Lorentz Force on an electric charge occurs when the charge moves through a magnetic field. This force is perpendicular to the direction of the charge and also perpendicular to the direction of the magnetic field. It is a vector combination of the two forces.