Question

An $\alpha -$particle travels at an angle of ${30}^{\circ }$ to a magnetic field of $0.8\text{T}$ with a velocity of ${10}^5\text{m/s}$. The magnitude of magnetic force will be -

• A. $1.28\times {10}^{-16}\text{N}$
• B. $1.28\times {10}^{-10}\text{N}$
• C. $1.28\times {10}^{-14}\text{N}$
• D. $1.28\times {10}^{-12}\text{N}$

Answer 1

The correct option is C $1.28\times {10}^{-14}\text{N}$

Force on a charged particle moving in a magnetic field $\overrightarrow{B}$ with velocity $\overrightarrow{v}$ is given by$$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$$$\Rightarrow F=qvB\sin \theta ...\left(1\right)$

Given:
Charge on $\alpha -$ particle, $q=+2e=+2\times 1.6\times {10}^{-19}\text{C}$

Velocity of charged particle, $v={10}^5\text{m/s}$

Magnetic field, $B=0.8\text{T}$

Angle between the velocity and the magnetic field, $\theta ={30}^{\circ }$

On substituting all the values in $\left(1\right)$, we get,

$F=qvB\sin \theta =2\times 1.6\times {10}^{-19}\times {10}^5\times 0.8\times \sin {30}^{\circ }$

$\therefore F=1.28\times {10}^{-14}\text{N}$

Hence, option $\left(C\right)$ is the correct answer.