A single circular loop of radius 1.00 , m car | vedclass

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(A) The torque $\vec{	au}$ on a current-carrying loop in a magnetic field is given by $\vec{	au} = \vec{\mu} 	imes \vec{B}$,where $\vec{\mu}$ is th

Vedclass · Accepted Answer

$1.57 	imes 10^{-2} \, N \cdot m$

Vedclass · Answer

(A) The torque $\vec{	au}$ on a current-carrying loop in a magnetic field is given by $\vec{	au} = \vec{\mu} 	imes \vec{B}$,where $\vec{\mu}$ is the magnetic dipole moment and $\vec{B}$ is the magnetic field.The magnitude of the torque is $	au = \mu B \sin 	heta$,where $	heta$ is the angle between the normal to the loop and the magnetic field.Since the magnetic field is parallel to the plane of the loop,the normal to the loop is perpendicular to the magnetic field. Thus,$	heta = 90^{\circ}$ and $\sin 90^{\circ} = 1$.The magnetic moment $\mu = I A = I (\pi r^2)$.Given: $I = 10.0 \, mA = 10.0 	imes 10^{-3} \, A$,$r = 1.00 \, m$,$B = 0.500 \, T$.$	au = (I \pi r^2) B = (10.0 	imes 10^{-3} \, A) 	imes (\pi 	imes (1.00 \, m)^2) 	imes (0.500 \, T)$$	au = 10.0 	imes 10^{-3} 	imes 3.14159 	imes 0.500 = 1.570795 	imes 10^{-2} \, N \cdot m \approx 1.57 	imes 10^{-2} \, N \cdot m$.

$A$ single circular loop of radius $1.00 \, m$ carries a current of $10.0 \, mA$. It is placed in a uniform magnetic field of magnitude $0.500 \, T$ that is directed parallel to the plane of the loop as shown in the figure. The magnitude of the torque exerted on the loop by the magnetic field is:

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