Question:
A circular coil connected to a battery of emf E produced a magnetic field at its centre. The coil is unwound, stretched to double its length and rewound into a coil of \( \left(\frac{1}{3}\right)^{\text{rd}} \) of its initial radius. If this coil is connected to a battery of emf E' to produce same magnetic field at its centre, then E' is:
A circular coil connected to a battery of emf E produced a magnetic field at its centre. The coil is unwound, stretched to double its length and rewound into a coil of \( \left(\frac{1}{3}\right)^{\text{rd}} \) of its initial radius. If this coil is connected to a battery of emf E' to produce same magnetic field at its centre, then E' is:
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Stretching a wire to \( k \) times its original length always increases its electrical resistance by a factor of \( k^2 \), since the length increases and the cross-sectional area decreases simultaneously.
Updated On: Jun 8, 2026
- \( \frac{2E}{9} \)
- \( \frac{3E}{7} \)
- \( \frac{9E}{4} \)
- \( \frac{E}{6} \)
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The Correct Option is A
Solution and Explanation
Concept:
The magnetic field \( B \) at the center of a circular loop coil carrying a current \( I \) with \( N \) turns is:
\[
B = \frac{\mu_0 N I}{2R}
\]
The total length of the wire is \( L = N \cdot (2\pi R) \). The resistance of the wire is \( R_w = \rho \frac{L}{A} \). When stretched to double its length (\( L' = 2L \)), its cross-sectional area halves, causing its total resistance to increase by a factor of 4: \( R_w' = 4R_w \).
Step 1: Finding the new number of turns \( N' \).
The new radius is given as \( R' = \frac{1}{3}R \). The total new stretched length is \( L' = 2L \): \[ N' \cdot (2\pi R') = 2 \cdot [N \cdot (2\pi R)] \implies N' \left(\frac{1}{3}R\right) = 2NR \implies N' = 6N \]
Step 2: Setting up the magnetic field equation.
We want the new magnetic field \( B' \) to equal the original field \( B \): \[ B' = B \implies \frac{\mu_0 N' I'}{2R'} = \frac{\mu_0 N I}{2R} \implies \frac{(6N) I'}{\frac{1}{3}R} = \frac{N I}{R} \implies 18 I' = I \implies I' = \frac{1}{18}I \]
Step 3: Relating current to battery EMF to solve for E'.
Using Ohm's Law \( I = \frac{E}{R_w} \) and noting that the new resistance is \( R_w' = 4R_w \): \[ I' = \frac{E'}{4R_w} \] Substitute this into our current relationship: \[ \frac{E'}{4R_w} = \frac{1}{18}\left(\frac{E}{R_w}\right) \implies E' = \frac{4}{18}E = \frac{2}{9}E \]
Step 1: Finding the new number of turns \( N' \).
The new radius is given as \( R' = \frac{1}{3}R \). The total new stretched length is \( L' = 2L \): \[ N' \cdot (2\pi R') = 2 \cdot [N \cdot (2\pi R)] \implies N' \left(\frac{1}{3}R\right) = 2NR \implies N' = 6N \]
Step 2: Setting up the magnetic field equation.
We want the new magnetic field \( B' \) to equal the original field \( B \): \[ B' = B \implies \frac{\mu_0 N' I'}{2R'} = \frac{\mu_0 N I}{2R} \implies \frac{(6N) I'}{\frac{1}{3}R} = \frac{N I}{R} \implies 18 I' = I \implies I' = \frac{1}{18}I \]
Step 3: Relating current to battery EMF to solve for E'.
Using Ohm's Law \( I = \frac{E}{R_w} \) and noting that the new resistance is \( R_w' = 4R_w \): \[ I' = \frac{E'}{4R_w} \] Substitute this into our current relationship: \[ \frac{E'}{4R_w} = \frac{1}{18}\left(\frac{E}{R_w}\right) \implies E' = \frac{4}{18}E = \frac{2}{9}E \]
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