Question

A circular loop of radius $a$ is carrying a current $i$ and is placed in a two dimensional magnetic field. The centre of loop coincides with the centre of field. The strength of magnetic field at the periphery of loop is $B$. The magnetic force experienced by wire is

• A. $\pi iaB$
• B. $4\pi iaB$
• C. Zero
• D. $2\pi iaB$

Answer 1

The correct option is D $2\pi iaB$

The direction of magnetic field is radially outwards at each point on the circumference of current carrying wire and it has a magnitude $B$ at that peripheral position.


Using the relation for magnetic force experienced by a small current element on the circumference of wire,

$\overrightarrow{dF}=i(\overrightarrow{dl}\times \overrightarrow{B})$

$\overrightarrow{dF}=idlB\sin {90}^{\circ }=Bidl\hat{k}$
$(\because$ angle between $\overrightarrow{dl}$ and $\overrightarrow{B}$ is ${90}^{\circ }$ as $\overrightarrow{B}$ is radially outwards)

$dF=iBdl$
Here $B$ & $i$ are constants.

Thus integrating for the closed loop we get,

$\int dF=F_{net}$

or, $F_{net}=Bi\int dl=Bi\left(2\pi a\right)$

$[\because \int dl$ = circumference of loop]

$\Rightarrow F_{net}=2\pi iaB$

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