JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A closely wounded circular coil of radius \(5\,cm\) produces a magnetic field of \(37.68 \times 10^{-4}\,T\) at its center. The current through the coil is \(......A\). [Given, number of turns in the coil is \(100\) and \(\pi=3.14]\)
- A \(3\)
- B \(6\)
- C \(9\)
- D \(12\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(B _{\text {centre}}=\frac{N \mu_{0} I}{2 R }\) \(37.68 \times 10^{-4}=\frac{100 \times 4 \pi \times 10^{-7} \times I }{2 \times 5 \times 10^{-2}}\) \(I =3\,A\)
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