Question

Q1.A particle of charge 2μC is at rest in a magnetic field B = -3K T. Magnetic lorentz force on the charge particle with respect to an observer moving with velocity V=4j m/s will be

12x10-6 iN

24x10-6 iN

6x10-6 KN

Zero​

Answer 1

Step-by-step explanation:

Given:

A particle of charge 2μC is at rest in a magnetic field B = -3k T.

To find:

Magnetic lorentz force on the charge particle with respect to an observer moving with velocity V=4j m/s?

Calculation:

First of all, calculate the velocity of the charge with respect to the observer :

$\rm \vec{v}_{co} = \vec{v}_{c} - \vec{v}_{o} </p><p>$

$\rm \implies\vec{v}_{co} = 0 - 4 \hat{j} </p><p>$

$\rm \implies\vec{v}_{co} = - 4 \hat{j} </p><p>$

Now, Lorentz Force is given as :

$\rm \: \vec{F} = q( \vec{v}_{co} \times \vec{B}) </p><p>$

$\rm \implies \: \vec{F} = q( - 4 \hat{j} \times - 3 \hat{k})$

$\rm \implies \: \vec{F} = 2 \times {10}^{ - 6} ( - 4 \hat{j} \times - 3 \hat{k})$

$\rm \implies \: \vec{F} = 24 \times {10}^{ - 6} \: \hat{i}$

So, force is 24 × 10^(-6) i Newton.