A current of 0.1, A circulates around a coil | 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:$B = \frac{\mu_0 N i}{2r}$Given values:$i = 0.1 \,

Vedclass · Accepted Answer

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

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(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns is given by the formula:$B = \frac{\mu_0 N i}{2r}$Given values:$i = 0.1 \, A$$N = 100$$r = 5 \, cm = 5 	imes 10^{-2} \, m$$\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$Substituting these values into the formula:$B = \frac{(4\pi 	imes 10^{-7}) 	imes 100 	imes 0.1}{2 	imes 5 	imes 10^{-2}}$$B = \frac{4\pi 	imes 10^{-7} 	imes 10}{10 	imes 10^{-2}}$$B = \frac{4\pi 	imes 10^{-6}}{10^{-1}}$$B = 4\pi 	imes 10^{-5} \, T$

$A$ current of $0.1\, A$ circulates around a coil of $100$ turns and having a radius equal to $5\, cm$. The magnetic field set up at the centre of the coil is $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$

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