Question

The figure shows a circular wire-loop of radius $a=20\text{cm}$, carrying a current $I$, placed in a perpendicular magnetic field $4\text{T}$. If the tension in the wire is $4\text{N}$, find the value of $I$.

• A. $0.5\text{A}$
• B. $1\text{A}$
• C. $5\text{A}$
• D. $10\text{A}$

Answer 1

The correct option is C $5\text{A}$

Free body digram:


If we take any small segment of the loop,

Force due to magnetic field
$d{F}_B=I(\overrightarrow{dl}\times \overrightarrow{B})$

$\Rightarrow d{F}_B=I\left(dl\right)B\sin {90}^{\circ }=I\left(dl\right)B$

$($direction shown in diagram$)$

If $T$ is the tension in the wire, from figure, we can see that

$2T\sin \theta =I\left(dl\right)B$

$($as all the forces are balanced$)$

For small values of $\theta ,\sin \theta \simeq \theta$

$\Rightarrow 2T\theta =IdlB$

$\Rightarrow 2T\theta =I\left(2\theta a\right)B[\because dl=a(2\theta \left)\right]$

$\Rightarrow T=IaB$

Given,

$B=4\text{T}$, $a=20\text{cm}$, $T=4\text{N}$

$\Rightarrow 4=I\times 20\times {10}^{-2}\times 4$

$\therefore I=5\text{A}$

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