JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A straight wire carrying a current of \(14\,A\) is bent into a semicircular are of radius \(2.2\,cm\) as shown in the figure. The magnetic field produced by the current at the centre \((O)\) of the arc. is \(.........\,\times 10^{-4}\, T\)
- A \(4\)
- B \(6\)
- C \(2\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(B _{\text {at } 0}=\frac{\mu_0 I }{4 R }=\frac{4 \pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}\) \(=2 \times 10^{-4}\,T\)
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