(C) The torque $\tau$ acting on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$,where $\theta$ is the angle betwe

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(C) The torque $\tau$ acting on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$,where $\theta$ is the angle between the normal to the coil and the magnetic field.Given:Number of turns $N = 30$Radius $r = 8.0\, cm = 0.08\, m$Current $I = 6.0\, A$Magnetic field $B = 1.0\, T$Angle $\theta = 60^o$Area $A = \pi r^2 = 3.14 \times (0.08)^2 = 3.14 \times 0.0064 = 0.020096\, m^2$Substituting the values:$\tau = 30 \times 6.0 \times 0.020096 \times 1.0 \times \sin 60^o$$\tau = 180 \times 0.020096 \times \frac{\sqrt{3}}{2}$$\tau = 90 \times 0.020096 \times 1.732$$\tau \approx 3.133\, Nm$Rounding to two significant figures,the magnitude of the counter torque is $3.1\, Nm$.

Class 12 Physics Moving Charges and Magnetism Torque , Potential Energy and Work Done in Mangetic Field

Medium

$A$ circular coil of $30$ turns and radius $8.0\, cm$ carrying a current of $6.0\, A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0\, T$. The field lines make an angle of $60^o$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (in $, Nm$)

A
$4$
B
$6$
C
$3.1$
D
$2.8$

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Moving Charges and Magnetism

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Torque , Potential Energy and Work Done in Mangetic Field

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$A$ rectangular coil has $2n$ turns,an area $2a$ and carries a current $2I$. The plane of the coil is at $60^{\circ}$ to a horizontal uniform magnetic field of flux density $B$. The torque on the coil due to magnetic force is

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The torque required to hold a small circular coil of $10$ turns, area $1 \,mm^{2}$ and carrying a current of $\left(\frac{21}{44}\right) \,A$ in the middle of a long solenoid of $10^{3} \,turns/m$ carrying a current of $2.5 \,A$, with its axis perpendicular to the axis of the solenoid is

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