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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2023
  • Easy
  • Numerical

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 \(.........\,\times 10^{-4}\, T\)

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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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