Question

A conductor in the form of a right angle $ABC$ with $AB=3\text{cm}$ and $BC=4\text{cm}$ carries a current $10\text{A}$. There exists a uniform magnetic field of magnitude $5\text{T}$ perpendicular to the plane of conductor. The magnetic force acting on conductor will be: (Assume the conductor is in gravity free space)

• A. $1.5\text{N}$
• B. $2\text{N}$
• C. $2.5\text{N}$
• D. $3.5\text{N}$

Answer 1

The correct option is C $2.5\text{N}$

The given conductor $\text{ABC}$ can be represented as,


On the part $\text{BC}$ of conductor:

$\overrightarrow{L}=-l_2\hat{i}=-(4\times {10}^{-2})\hat{i}\text{m}$
$\overrightarrow{B}=-5\hat{k}\text{T}$

Magnetic force:

$\overrightarrow{F_1}=i(\overrightarrow{L}\times \overrightarrow{B})=10\times [-(4\times {10}^{-2})\hat{i}\times (-5\left)\hat{k}\right]$

$\Rightarrow \overrightarrow{F_1}=200\times {10}^{-2}(-\hat{j})=-2\hat{j}\text{N}$

Similarly, on part $\text{AB}$ of the conductor:

$\overrightarrow{L}=+(3\times {10}^{-2})\hat{j}\text{m}$ and $\overrightarrow{B}=-5\hat{k}\text{T}$

Force acting on part $\text{AB}$:

$\overrightarrow{F_2}=i(\overrightarrow{L}\times \overrightarrow{B})=10\times \left[\right(3\times {10}^{-2})\hat{j}\times (-5\hat{k}]$

$\Rightarrow \overrightarrow{F_2}=-150\times {10}^{-2}\hat{i}=-1.5\hat{i}\text{N}$

Thus, the net force acting on wire $\text{ABC}$ is:

$|{\overrightarrow{F}}_{net}|=\sqrt{F_1^2+F_2^2}=\sqrt{{(-1.5)}^2+{(-2)}^2}$

$|{\overrightarrow{F}}_{net}|=\sqrt{6.25}=2.5\text{N}$

Hence, option (c) is the correct answer.