JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 \(\pi=3.14\) )

- A \(689\)
- B \(546\)
- C \(487\)
- D \(628\)
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…
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