A circular loop of radius $r$ having $N$ number of turns carries current $I$ is placed in a uniform magnetic field $\vec{B}$ parallel to the plane of the loop. The torque on the loop is

$N l π r^{2} B$
$N^{2} l π r^{2} B$
$N l^{2} π r^{2} B$
$N l π r^{2} B^{2}$

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Solution

The correct option is A
$N l π r^{2} B$

Step 1 : Given data
Radius of circular loop $= r$
Magnetic field $=$ $\vec{B}$
Step 2 : Formula used
As we know, the magnetic dipole moment is given by
$|\vec{M}| = N I A$
Where $N is number of turn, I is current, A is area of crossection$
$|\vec{M}| = N I π r^{2}$
$\vec{ζ} = \vec{M} \times \vec{B} w h e r e \vec{ζ} i s t o r q u e s .$
Where,
$M =$ magnetic moment
$B =$ magnetic field
Step 3 : To find the torque
$|\vec{ζ}| = M B \sin θ$
$M B = N I π r^{2}$
$|\vec{ζ}| = N I π r^{2} \sin 90^{0} \times B$
$|\vec{ζ}| = N I π r^{2} B$
Hence torque on the loop $N I π r^{2} B$
Hence, The correct answer is Option A

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A circular loop of radius r having Nnumber of turns carries current I is placed in a uniform magnetic field B→ parallel to the plane of the loop. The torque on the loop isNlπr2BN2lπr2BNl2πr2BNlπr2B2

A circular loop of radius $r$ having $N$ number of turns carries current $I$ is placed in a uniform magnetic field $\vec{B}$ parallel to the plane of the loop. The torque on the loop is

$N l π r^{2} B$
$N^{2} l π r^{2} B$
$N l^{2} π r^{2} B$
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