JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Figure shows a current carrying square loop ABCD of edge length is ' \(a\) ' lying in a plane. If the resistance of the \(A B C\) part is \(r\) and that of \(A D C\) part is 2 r , then the magnitude of the resultant magnetic field at centre of the square loop is
- A \(\frac{3 \pi \mu_0 \mathrm{I}}{\sqrt{2} \mathrm{a}}\)
- B \(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}}\)
- C \(\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}\)
- D \(\frac{2 \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{BC}}+\overrightarrow{\mathrm{B}}_{\mathrm{CD}}+\overrightarrow{\mathrm{B}}_{\mathrm{DA}}\)…
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