A coil of 100 turns and area $2 imes 10^{- 2}m^{2},\ $pivoted about a vertical diameter in a uniform magnetic field carries a current of $5A$. When the coil is held with its plane in North-South direction, it experience a torque of $0.3\ Nm$. When the plane is in East-West direction the torque is $0.4\ Nm$. The value of magnetic induction is (Neglect earth’s magnetic field)

Question

A coil of 100 turns and area $2 	imes 10^{- 2}m^{2},\ $pivoted about a vertical diameter in a uniform magnetic field carries a current of $5A$. When the coil is held with its plane in North-South direction, it experience a torque of $0.3\ Nm$. When the plane is in East-West direction the torque is $0.4\ Nm$. The value of magnetic induction is (Neglect earth’s magnetic field)

Dr. Abubacker · Accepted Answer

(d) $N = 100$ turns, area $\pi r^{2} = 2 	imes 10^{- 2}m^{2},\ i = 5A$
In north-south, torque $= 0.3\ Nm = 	au_{1}$
When plane is in east-west direction, torque $= 0.4\ Nm = 	au_{2}$
$B = \frac{\mu_{0}Ni}{2r}$
$M = NiA = 100 	imes 5 	imes 2 	imes 10^{- 2} = 10$
$	au_{1} = MB\sin	heta;\ \ 	au_{2} = MB\cos	heta$
$	herefore\frac{	au_{1}}{	au_{2}} = 	an	heta$
$	herefore	an	heta = \frac{0.3}{0.4}$
$	au_{1}^{2} = M^{2}B^{2}\sin^{2}	heta;\ 	au_{2}^{2} = M^{2}B^{2}\cos^{2}	heta$
$	herefore	au_{1}^{2} + 	au_{2}^{2} = (MB)^{2}$
$	herefore(0.09 + 0.16) = 10^{2}\ B^{2}$
$B^{2} = \frac{0.25}{100} \Rightarrow B = \frac{0.5}{10} = 0.05\ T$

Explanation

Reviewed by Dr. Abubacker

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