A circular coil of radius R having N number of turns carries a steady current I. The magnetic induction at the centre of the coil is 0.1 tesla. If the number of turns is doubled and the radius is halved, which one of the following will be the correct value for the magnetic induction at the centre of the coil?
- A.0.05 tesla
- B.0.2 tesla
- C.0.4 tesla
- D.0.8 tesla
Correct answer
Explanation
The correct answer is C, 0.4 tesla. The field at the centre of a coil is proportional to the number of turns and inversely proportional to the radius, so doubling N and halving R makes it four times larger.
For a circular coil, the magnetic field at the centre is B = μ₀NI / 2R. Doubling the turns doubles B, and halving the radius doubles it again, so the new field is 2 × 2 × 0.1 = 0.4 tesla, with the current unchanged. This is why electromagnets use many tightly wound turns of small radius to get a strong field from a modest current. The tesla is the SI unit of magnetic field, named after Nikola Tesla.
A is wrong because 0.05 tesla would mean the field had halved, which happens only if R doubles or N halves. B is wrong because 0.2 tesla accounts for only one of the two changes. D is wrong because 0.8 tesla would need an eightfold increase, which these changes do not give.
Exam tip: B at the centre of a coil = μ₀NI/2R, so B ∝ N and B ∝ 1/R.
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