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. \( \frac{2E}{9} \)
• B. \( \frac{3E}{7} \)
• C. \( \frac{9E}{4} \)
• D. \( \frac{E}{6} \)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Write the field at the centre.
A circular coil with $N$ turns carrying current $I$ makes a field at its centre: \[ B = \frac{\mu_0 N I}{2R} \] We must keep this field the same after the coil is changed.
Step 2: Track the wire and its resistance.
When the same wire is stretched to double its length, its cross section halves. Since resistance is $\rho L/A$, doubling $L$ and halving $A$ makes resistance go up by $4$ times: $R_w' = 4R_w$.
Step 3: Find the new number of turns.
Total wire length is $N(2\pi R)$. After stretching the length doubles and the new radius is $R' = R/3$: \[ N'(2\pi R') = 2N(2\pi R) \;\Rightarrow\; N'\left(\frac{R}{3}\right) = 2NR \;\Rightarrow\; N' = 6N \]
Step 4: Set the fields equal.
We want the new field to match the old: \[ \frac{\mu_0 N' I'}{2R'} = \frac{\mu_0 N I}{2R} \;\Rightarrow\; \frac{6N\,I'}{R/3} = \frac{N I}{R} \]
Step 5: Solve for the new current.
\[ 18 I' = I \;\Rightarrow\; I' = \frac{I}{18} \]
Step 6: Bring in the emf.
Using $I = E/R_w$ and $R_w' = 4R_w$: \[ \frac{E'}{4R_w} = \frac{1}{18}\frac{E}{R_w} \;\Rightarrow\; E' = \frac{4}{18}E = \frac{2E}{9} \] \[ \boxed{E' = \frac{2E}{9}} \]