Question

Magnetic field $\overrightarrow{B}=2\times {10}^{-5}\text{T}$ exists in the space in vertically downward direction. A proton is projected northward at $3\times {10}^6\text{m}/\text{s}$. What is the force acting on it?

• A. $9.6\times {10}^{-18}\text{N}$ $\text{West}$
• B. $9.6\times {10}^{-18}\text{N}$ $\text{East}$
• C. $4.8\times {10}^{-18}\text{N}$ $\text{West}$
• D. $4.8\times {10}^{-18}\text{N}$ $\text{East}$

Answer 1

The correct option is A $9.6\times {10}^{-18}\text{N}$ $\text{West}$


As per question,

$\overrightarrow{B}=-(2\times {10}^{-5}\text{T})\hat{j}$
$\overrightarrow{v}=-(3\times {10}^6\text{m/s})\hat{k}$
$q=1.6\times {10}^{-19}\text{C}$

So, magnetic force,

${\overrightarrow{F}}_m=q\overrightarrow{v}\times \overrightarrow{B}$

$\Rightarrow {\overrightarrow{F}}_m=1.6\times {10}^{-19}[-(3\times {10}^6\text{m/s})\hat{k}\times -(2\times {10}^{-5}\text{T}\left)\hat{j}\right]$

$\Rightarrow {\overrightarrow{F}}_m=-(9.6\times {10}^{-18}\text{N})\hat{i}$

The magnetic force on the proton is $9.6\times {10}^{-18}\text{N}$ towards west.