Question:
Two circular coils \(A\) and \(B\) are made from the same wire but the radius of coil \(A\) is twice that of coil \(B\). What is the ratio of potential difference applied across them, so that the magnetic field at their centres is the same?
Two circular coils \(A\) and \(B\) are made from the same wire but the radius of coil \(A\) is twice that of coil \(B\). What is the ratio of potential difference applied across them, so that the magnetic field at their centres is the same?
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For a coil made of a given wire, the field at the centre depends on V and the radius. Express B in terms of V and r.
Updated On: Nov 12, 2026
- \(3:1\)
- \(4:1\)
- \(6:1\)
- \(2:1\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The field at the centre of a coil is \(B=\dfrac{\mu_0NI}{2r}\). Both coils use the same wire (same \(\rho\) and cross-section).
Step 2: Current in the Coil
Wire length is \(2\pi rN\), so \(R\propto Nr\) and \(I=\dfrac{V}{R}\propto\dfrac{V}{Nr}\).
Step 3: Field in Terms of V
\(B\propto\dfrac{N}{r}\cdot\dfrac{V}{Nr}=\dfrac{V}{r^{2}}\).
Step 4: Equal Fields
For equal \(B\), \(V\propto r^{2}\), so \(\dfrac{V_A}{V_B}=\left(\dfrac{2r}{r}\right)^{2}=4\).
Final Answer:
The ratio of potential differences is 4 to 1. \[ \boxed{\text{(B) }4:1} \]
The field at the centre of a coil is \(B=\dfrac{\mu_0NI}{2r}\). Both coils use the same wire (same \(\rho\) and cross-section).
Step 2: Current in the Coil
Wire length is \(2\pi rN\), so \(R\propto Nr\) and \(I=\dfrac{V}{R}\propto\dfrac{V}{Nr}\).
Step 3: Field in Terms of V
\(B\propto\dfrac{N}{r}\cdot\dfrac{V}{Nr}=\dfrac{V}{r^{2}}\).
Step 4: Equal Fields
For equal \(B\), \(V\propto r^{2}\), so \(\dfrac{V_A}{V_B}=\left(\dfrac{2r}{r}\right)^{2}=4\).
Final Answer:
The ratio of potential differences is 4 to 1. \[ \boxed{\text{(B) }4:1} \]
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