Question

A particle of mass $m=0.5g$ having charge of $q=1.3C$ enters a magnetic field $B=0.5Wb$ orthogonally, experiences a force $F_B=2.6N$. Then, $K.E.$ of the particle is (in $J$)

• A. $4$
• B. $2\times {10}^{-3}$
• C. $2$
• D. $4\times {10}^{-3}$

Answer 1

The correct option is D $4\times {10}^{-3}$

Given :

Mass of a particle, $m=0.5g=0.0005kg$.

Charge of a particle, $q=1.3C$.

Strength of magnetic field, $B=0.5Wb$.

Force experienced by the particle, $F_B=2.6N$.

Now, the magnetic force experienced by the particle in magnetic field is

$F_B=Bqv$

where, $v$ is the velocity of the particle.

Therefore,

$v=\frac{F_B}{Bq}$

$v=\frac{2.6}{\left(0.5\right)\left(1.3\right)}$

$v=4\frac{m}{s}$ ....................(1)

The kinetic energy, E of the particle is given by

$E=\frac{1}{2}m{v}^2$

$E=\left(0.5\right)\left(0.0005\right)(4{)}^2$

$E=\left(0.00025\right)\left(16\right)$

$E=0.004J$

$E=4\times {10}^{-3}J$