Two concentric circular coils of ten turns ea | vedclass

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(D) The magnetic field at the centre of a circular coil of $N$ turns carrying current $i$ with radius $r$ is given by $B = \frac{\mu_0 N i}{2r}$.Given

Vedclass · Accepted Answer

$\frac{5}{4} \mu_0$

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(D) The magnetic field at the centre of a circular coil of $N$ turns carrying current $i$ with radius $r$ is given by $B = \frac{\mu_0 N i}{2r}$.Given: $N = 10$,$r_1 = 0.2 \ m$,$i_1 = 0.2 \ A$,$r_2 = 0.4 \ m$,$i_2 = 0.3 \ A$.Since the currents are in opposite directions,the net magnetic field at the centre is the difference between the two fields:$B_{net} = |B_1 - B_2| = |\frac{\mu_0 N i_1}{2r_1} - \frac{\mu_0 N i_2}{2r_2}|$$B_{net} = \frac{\mu_0 	imes 10}{2} |\frac{0.2}{0.2} - \frac{0.3}{0.4}|$$B_{net} = 5 \mu_0 |1 - 0.75|$$B_{net} = 5 \mu_0 	imes 0.25 = 1.25 \mu_0 = \frac{5}{4} \mu_0 \ Wb/m^2$.

Two concentric circular coils of ten turns each are situated in the same plane. Their radii are $20 \ cm$ and $40 \ cm$ and they carry respectively $0.2 \ A$ and $0.3 \ A$ current in opposite directions. The magnetic field in $Wb/m^2$ at the centre is:

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