NEET · Physics · STD 12 - 4. Moving charges and magnetism
A tightly wound \(100\) turns coil of radius \(10 \mathrm{~cm}\) carries a current of \(7 \mathrm{~A}\). The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as \(4 \pi \times 10^{-7} \mathrm{SI}\) units):
- A \(4.4 \mathrm{~T}\)
- B \(4.4 \mathrm{mT}\)
- C \(44 \mathrm{~T}\)
- D \(44 \mathrm{mT}\)
Answer & Solution
Correct Answer
(B) \(4.4 \mathrm{mT}\)
Step-by-step Solution
Detailed explanation
The magnitude of magnetic field due to circular coil of \(N\) turns is given by \(B_c=\frac{\mu_0 i N}{2 R} \)
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