A coil of radius 5 cm has 98 closely wound turns of a metal wire. Magnetic field of strength \( 6.2 \times 10^{-3}\,\mathrm{T} \) is produced at the centre of this coil, when a current of 5.2 A is passed through it. Magnetic field of same strength is to be produced at the centre of another coil with 10 cm radius. Select the correct possible way to do so.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Current of 10.4 A should be passed through the second coil of 98 turns.
The magnetic field at the centre of a circular coil of \( N \) turns and radius \( r \) is
\( \displaystyle B = \frac{\mu_0 N I}{2r} \;\Rightarrow\; B \propto \frac{N I}{r} \)
The radius is doubled (5 cm to 10 cm), so the field stays the same only if the product \( N I \) is also doubled. For the first coil \( N I = 98 \times 5.2 \); the second coil needs \( N I = 2 \times 98 \times 5.2 \). Checking \( N I \) for each choice of the second coil:
98 turns, 10.4 A: \( N I = 98 \times 10.4 = 2 \times 98 \times 5.2 \), exactly double, so \( B \) is unchanged.
196 turns, 10.4 A: \( N I = 196 \times 10.4 = 4 \times 98 \times 5.2 \), giving \( 2B \).
49 turns, 5.2 A: \( N I = 49 \times 5.2 \), half the original, giving \( B/4 \).
196 turns, 2.6 A: \( N I = 196 \times 2.6 = 98 \times 5.2 \), the same as before, giving only \( B/2 \).