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(C) Let the tension developed in the wire be $T$ and the radius of the circular loop be $r$. Consider a small element of the wire of length $dl$ subte

Vedclass · Accepted Answer

$0.35$

Vedclass · Answer

(C) Let the tension developed in the wire be $T$ and the radius of the circular loop be $r$. Consider a small element of the wire of length $dl$ subtending an angle $d	heta$ at the center. The magnetic force acting on this element is $dF = B I dl = B I (r d	heta)$. This force is directed radially outward. The tension $T$ at the ends of this element provides a restoring force directed radially inward. The component of tension providing the inward force is $2 T \sin(d	heta / 2) \approx 2 T (d	heta / 2) = T d	heta$. Equating the forces for equilibrium: $T d	heta = B I r d	heta$, which gives $T = B I r$. Since the total length of the wire is $l = 2 \pi r$, we have $r = l / (2 \pi)$. Substituting $r$ in the tension formula: $T = B I (l / 2 \pi)$. Given $B = 2.0 \,T$, $I = 1.1 \,A$, and $l = 1 \,m$: $T = (2.0 	imes 1.1 	imes 1) / (2 	imes 3.14) = 2.2 / 6.28 \approx 0.35 \,N$.

$A$ loop of flexible conducting wire lies in a magnetic field of $2.0 \,T$ with its plane perpendicular to the field. The length of the wire is $1 \,m$. When a current of $1.1 \,A$ is passed through the loop, it opens into a circle. The tension developed in the wire is (in $\,N$)

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