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A small circular loop of radius 0.3 cm lies parallel to a much larger circular loop of radius 20 cm. The centre of the small loop sits on the axis of the large loop, 15 cm from its centre. If a current of 2.0 A flows in the small loop, what is the magnetic flux linked with the large loop?

1. 9.1*10⁻¹¹ weber
2. 6*10⁻¹¹ weber
3. 3.3*10⁻¹¹ weber
4. 6.6*10⁻⁹ weber

Correct answer: 9.1*10⁻¹¹ weber

Solution

By reciprocity, the mutual inductance is M = mu0 * pi * R² * a² / (2 (R² + x²)^(3/2)), where R = 0.20 m (large loop), a = 0.003 m (small loop), x = 0.15 m. The field of the small loop is awkward off-axis, so compute the flux through the big loop as M*I where M uses the on-axis field of the big loop acting over the small loop's area (reciprocity). Plugging in gives flux ~ 9.1*10⁻¹¹ Wb.

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