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(B) The length of the wire is $L = 10 \ m$. When it is bent into a circular loop of radius $r$,the circumference is $2 \pi r = L = 10 \ m$. Thus,$r =

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$50 	imes 10^{-4} \ N \ m$

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(B) The length of the wire is $L = 10 \ m$. When it is bent into a circular loop of radius $r$,the circumference is $2 \pi r = L = 10 \ m$. Thus,$r = \frac{10}{2 \pi} \ m$. The area of the loop is $A = \pi r^2 = \pi \left( \frac{10}{2 \pi} ight)^2 = \pi \left( \frac{100}{4 \pi^2} ight) = \frac{25}{\pi} \ m^2$. The magnetic moment of the loop is $M = I 	imes A = 1 	imes \frac{25}{\pi} = \frac{25}{\pi} \ A \ m^2$. The maximum torque acting on the loop in a magnetic field $B$ is given by $	au_{max} = M 	imes B$. Substituting the values,$	au_{max} = \left( \frac{25}{\pi} ight) 	imes (2 \pi 	imes 10^{-4}) = 50 	imes 10^{-4} \ N \ m$.

$A$ wire of length $10 \ m$ carrying a current of $1 \ A$ is bent into a circular loop. If a magnetic field of $2 \pi \times 10^{-4} \ T$ is applied on the loop,then the maximum torque acting on it is

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