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(B) Number of turns on the circular coil,$n = 100$.Radius of each turn,$r = 8.0 \; cm = 0.08 \; m$.Current flowing in the coil,$I = 0.40 \; A$.The mag

Vedclass · Accepted Answer

$3.14 	imes 10^{-4} \; T$

Vedclass · Answer

(B) Number of turns on the circular coil,$n = 100$.Radius of each turn,$r = 8.0 \; cm = 0.08 \; m$.Current flowing in the coil,$I = 0.40 \; A$.The magnitude of the magnetic field at the centre of the coil is given by the formula:$B = \frac{\mu_0 n I}{2r}$Substituting the values:$B = \frac{(4\pi 	imes 10^{-7} \; T \cdot m/A) 	imes 100 	imes 0.40}{2 	imes 0.08 \; m}$$B = \frac{4\pi 	imes 10^{-7} 	imes 40}{0.16}$$B = \frac{160\pi 	imes 10^{-7}}{0.16} = 1000\pi 	imes 10^{-7} = \pi 	imes 10^{-4} \; T$Using $\pi \approx 3.14$,we get:$B = 3.14 	imes 10^{-4} \; T$.

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \; cm$,carries a current of $0.40 \; A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

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