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

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(D) The magnetic field at the centre of a circular coil is given by the formula: $B = N \frac{\mu_0 I}{2r}$.Given: $N = 50$,$r = 4\,cm = 4 	imes 10^{

Vedclass · Accepted Answer

$1.57$

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(D) The magnetic field at the centre of a circular coil is given by the formula: $B = N \frac{\mu_0 I}{2r}$.Given: $N = 50$,$r = 4\,cm = 4 	imes 10^{-2}\,m$,$I = 2\,A$,and $\mu_0 = 4\pi 	imes 10^{-7}\,T\cdot m/A$.Substituting the values:$B = 50 	imes \frac{4\pi 	imes 10^{-7} 	imes 2}{2 	imes 4 	imes 10^{-2}}$$B = \frac{50 	imes 8\pi 	imes 10^{-7}}{8 	imes 10^{-2}}$$B = 50\pi 	imes 10^{-5} = 5\pi 	imes 10^{-4}\,T$.Using $\pi \approx 3.14$,$B \approx 15.7 	imes 10^{-4}\,T = 1.57 	imes 10^{-3}\,T$.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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