A conducting ring of mass 2 kg and radius 0. | vedclass

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Question

(A) The magnetic moment of the current-carrying ring is $\mu = i A = i (\pi r^2)$.The torque acting on the ring due to the magnetic field is given by

Vedclass · Accepted Answer

$40 \pi \ rad/s^2$

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(A) The magnetic moment of the current-carrying ring is $\mu = i A = i (\pi r^2)$.The torque acting on the ring due to the magnetic field is given by $\vec{	au} = \vec{\mu} 	imes \vec{B}$.Since the magnetic field is in the plane of the ring,the angle between the magnetic moment vector (perpendicular to the plane of the ring) and the magnetic field vector is $90^\circ$.Therefore,the magnitude of the torque is $	au = \mu B \sin(90^\circ) = i \pi r^2 B$.The ring will rotate about its vertical diameter. The moment of inertia of a ring about its diameter is $I = \frac{1}{2} m r^2$.Using the rotational analog of Newton's second law,$	au = I \alpha$,we have:$i \pi r^2 B = \left( \frac{1}{2} m r^2 ight) \alpha$$\alpha = \frac{2 i \pi B}{m}$Substituting the given values ($i = 4 \ A$,$B = 10 \ T$,$m = 2 \ kg$):$\alpha = \frac{2 	imes 4 	imes \pi 	imes 10}{2} = 40 \pi \ rad/s^2$.

$A$ conducting ring of mass $2 \ kg$ and radius $0.5 \ m$ is placed on a smooth horizontal plane. The ring carries a current $i = 4 \ A$. $A$ horizontal magnetic field $B = 10 \ T$ is switched on at time $t = 0$ as shown in the figure. The initial angular acceleration of the ring will be:

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