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

Two identical circular wires of radius \(20\,cm\) and carrying current \(\sqrt{2}\,A\) are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wire is \(.............\times 10^{-8}\,T\). (Take \(\pi=3.14\) )

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Answer & Solution

Correct Answer

(D) \(628\)

Step-by-step Solution

Detailed explanation

\(\text { Magnetic field } B_C \text { at center }=\frac{\mu_0 i }{2 r }\)

\(=\frac{4 \pi \times 10^{-7}}{2 \times 0.2} \times \sqrt{2}\,T\)

Net magnetic field is

\(B _{ C } \sqrt{2}= \frac{4 \pi \times 10^{-7} \times \sqrt{2}}{2 \times 0.2} \times \sqrt{2} T =2 \pi \times 10^{-6}\,T\)

\(=200 \pi \times 10^{-8}\,T\)

\(=2 \times 314 \times 10^{-8}\,T\)

\(=628 \times 10^{-8}\,T\)

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