Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 2.0 A . The coils have 50 and 100 turns respectively and are positioned so that their planes and centres coincide. Determine the magnitude of the

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Correct answer: A. 8π × 10^ -4 T

The magnetic field at the centre of a circular coil is given by

B = \dfrac{\mu_0 N I}{2r}

For the first coil:

B_1 = \dfrac{\mu_0 N_1 I_1}{2r_1} = \dfrac{4\pi \times 10^{-7} \times 50 \times 2.0}{2 \times 0.05} = 4\pi \times 10^{-4} \text{ T}

For the second coil:

B_2 = \dfrac{\mu_0 N_2 I_2}{2r_2} = \dfrac{4\pi \times 10^{-7} \times 100 \times 2.0}{2 \times 0.10} = 4\pi \times 10^{-4} \text{ T}

Since the currents flow in the same sense, the magnetic fields are in the same direction. The net magnetic field is:

B = B_1 + B_2 = 4\pi \times 10^{-4} + 4\pi \times 10^{-4} = 8\pi \times 10^{-4} \text{ T}

Two circular coils of radii 5.0 cm and 10 cm carry equal currents...

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