Two concentric circular coils of 10 turns eac | vedclass

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(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 n I}{2r}$.Since the currents are in opposite directions, the net

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$5 \pi 	imes 10^{-7} \, T$

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(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 n I}{2r}$.Since the currents are in opposite directions, the net magnetic field is the difference between the fields produced by each coil: $B_{	ext{net}} = |B_1 - B_2|$.Given: $n_1 = n_2 = 10$, $r_1 = 0.2 \, m$, $r_2 = 0.4 \, m$, $I_1 = 0.2 \, A$, $I_2 = 0.3 \, A$.$B_1 = \frac{\mu_0 	imes 10 	imes 0.2}{2 	imes 0.2} = \frac{10 \mu_0}{2} = 5 \mu_0$.$B_2 = \frac{\mu_0 	imes 10 	imes 0.3}{2 	imes 0.4} = \frac{3 \mu_0}{0.8} = 3.75 \mu_0$.$B_{	ext{net}} = 5 \mu_0 - 3.75 \mu_0 = 1.25 \mu_0$.Substituting $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$:$B_{	ext{net}} = 1.25 	imes 4 \pi 	imes 10^{-7} = 5 \pi 	imes 10^{-7} \, T$.

Two concentric circular coils of $10$ 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 at the centre is $(\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A)$

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