A coil of 50 turns and 4 cm radius carries 2 | vedclass

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(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns is given by the formula: $B = N \left( \frac{\mu_0 I}{2r} \right)$.Given: $

Vedclass · Accepted Answer

$1.57$

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(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns is given by the formula: $B = N \left( \frac{\mu_0 I}{2r} ight)$.Given: $N = 50$,$I = 2$ $A$,$r = 4$ cm = $0.04$ m = $4 	imes 10^{-2}$ m,and $\mu_0 = 4\pi 	imes 10^{-7}$ $T$ m/$A$.Substituting the values:$B = 50 	imes \frac{4\pi 	imes 10^{-7} 	imes 2}{2 	imes 4 	imes 10^{-2}}$$B = 50 	imes \frac{8\pi 	imes 10^{-7}}{8 	imes 10^{-2}}$$B = 50 	imes \pi 	imes 10^{-5}$ $T$$B = 50 	imes 3.14159 	imes 10^{-5}$ $T$$B = 157.08 	imes 10^{-5}$ $T$ = $1.5708 	imes 10^{-3}$ $T$.Since $1$ mT = $10^{-3}$ $T$,the magnetic field is $1.57$ mT.

$A$ coil of $50$ turns and $4$ cm radius carries $2$ $A$ current. The magnetic field at its centre is ...... mT.

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