Question:
Two planar concentric rings of metal wire having radii '\(r_1\)' and '\(r_2\)' (with \(r_1 > r_2\)) are placed in air. The current 'I' is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( \(μ_0\) = permeability of free space)
Two planar concentric rings of metal wire having radii '\(r_1\)' and '\(r_2\)' (with \(r_1 > r_2\)) are placed in air. The current 'I' is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( \(μ_0\) = permeability of free space)
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A rotating charge is equivalent to a current q f, and a circular loop gives mu_0 I over 2R at the centre.
Updated On: Oct 1, 2026
- \(\frac{μ_0π\,r_1^2}{2\,r_2}\)
- \(\frac{μ_0π\,r_2^2}{2\,r_1}\)
- \(\frac{μ_0π\,(r_1+r_2)^2}{2\,r_1}\)
- \(\frac{μ_0π\,(r_1-r_2)^2}{2\,r_2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Equivalent current:
The ring completes f revolutions per second, so charge q passes any point f times each second. The equivalent current is \(I = qf\).
Step 2: Field at the centre:
\[ B = \frac{\mu_0I}{2R} = \frac{\mu_0qf}{2R} \]
Options with f in the denominator are wrong, since a faster rotation gives a larger current and a larger field.
Final Answer:
The magnetic induction is \(\frac{\mu_0qf}{2R}\), option (C). \[ \boxed{\frac{\mu_0 qf}{2R}} \]
The ring completes f revolutions per second, so charge q passes any point f times each second. The equivalent current is \(I = qf\).
Step 2: Field at the centre:
\[ B = \frac{\mu_0I}{2R} = \frac{\mu_0qf}{2R} \]
Options with f in the denominator are wrong, since a faster rotation gives a larger current and a larger field.
Final Answer:
The magnetic induction is \(\frac{\mu_0qf}{2R}\), option (C). \[ \boxed{\frac{\mu_0 qf}{2R}} \]
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