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 \(\pi=3.14\) )

Pick an option to check your answer.
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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