Two concentric coils X and Y of radii 16 , cm | vedclass

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(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2R}$.For coil $X$: $N_X = 20$,$I_X = 16 \, A$,$R_X = 0.16 \,

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$5\pi 	imes 10^{-4} \, T$ towards east

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(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2R}$.For coil $X$: $N_X = 20$,$I_X = 16 \, A$,$R_X = 0.16 \, m$.$B_X = \frac{4\pi 	imes 10^{-7} 	imes 20 	imes 16}{2 	imes 0.16} = \frac{4\pi 	imes 10^{-7} 	imes 320}{0.32} = 4\pi 	imes 10^{-4} \, T$.Since the current is anticlockwise for an observer facing west,by the right-hand rule,the field points towards the west.For coil $Y$: $N_Y = 25$,$I_Y = 18 \, A$,$R_Y = 0.10 \, m$.$B_Y = \frac{4\pi 	imes 10^{-7} 	imes 25 	imes 18}{2 	imes 0.10} = \frac{4\pi 	imes 10^{-7} 	imes 450}{0.20} = 9\pi 	imes 10^{-4} \, T$.Since the current is clockwise for an observer facing west,the field points towards the east.The net magnetic field $B_{net} = |B_Y - B_X| = |9\pi - 4\pi| 	imes 10^{-4} = 5\pi 	imes 10^{-4} \, T$.Since $B_Y > B_X$,the direction is the same as $B_Y$,which is towards the east.

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