A 100 turns coil shown in the figure carries | vedclass

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(B) The torque $	au$ acting on a current-carrying coil in a magnetic field is given by $	au = NBiA \sin 	heta$. Here,$N = 100$,$B = 0.2\, Wb/m^2$,$

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$0.32\, Nm$ tending to rotate the side $AD$ into the page

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(B) The torque $	au$ acting on a current-carrying coil in a magnetic field is given by $	au = NBiA \sin 	heta$. Here,$N = 100$,$B = 0.2\, Wb/m^2$,$i = 2\, A$,and the area $A = 8\, cm 	imes 10\, cm = 0.08\, m 	imes 0.1\, m = 0.008\, m^2$. Since the plane of the coil is parallel to the magnetic field lines,$	heta = 90^\circ$,so $\sin 90^\circ = 1$. Substituting the values: $	au = 100 	imes 0.2 	imes 2 	imes 0.008 = 0.32\, Nm$. Using Fleming's left-hand rule on side $AD$,the current flows downwards. The magnetic field points from $N$ to $S$ (left to right). The force $F = i(L 	imes B)$ on side $AD$ points into the page. Thus,the torque tends to rotate side $AD$ into the page.

$A$ $100$ turns coil shown in the figure carries a current of $2\, A$ in a magnetic field $B = 0.2\, Wb/m^2$. The torque acting on the coil is

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