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(D) Consider a small element of the ring of length $dl = r \, d	heta$ at an angle $	heta$. The magnetic force on this element is $dF = I \, dl \, B

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$10$

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(D) Consider a small element of the ring of length $dl = r \, d	heta$ at an angle $	heta$. The magnetic force on this element is $dF = I \, dl \, B = I \, r \, d	heta \, B$,acting radially outwards.To find the tension $T$,consider a semi-circular part of the ring. The total outward force on the semi-circle is the integral of the radial components of the force on each element.The net outward force $F_{net}$ on the semi-circle is given by $F_{net} = \int_{0}^{\pi} (I \, r \, B \, d	heta) \sin 	heta = I \, r \, B [-\cos 	heta]_{0}^{\pi} = 2 \, I \, r \, B$.This outward force is balanced by the tension $T$ acting at the two ends of the semi-circle. Thus,$2T = F_{net} = 2 \, I \, r \, B$.Therefore,$T = I \, r \, B$.Substituting the given values: $T = 100 \, A 	imes 0.5 \, m 	imes 0.2 \, T = 10 \, N$.

$A$ metal ring of radius $r = 0.5 \, m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2 \, T$ carries a current $I = 100 \, A$. The tension in newtons developed in the ring is:

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