MHT CET · Physics · Magnetic Effects of Current
Two identical current carrying coils with same centre are placed with their planes perpendicular to each other. If current \(I=\sqrt{2} \mathrm{~A}\) and radius of the coil is \(\mathrm{R}=1 \mathrm{~m}\), then magnetic field at centre is equal to ( \(\mu_0=\) permeability of free space)
- A \(\mu_0\)
- B \(\frac{\mu_0}{2}\)
- C \(2 \mu_0\)
- D \(\sqrt{2} \mu_0\)
Answer & Solution
Correct Answer
(A) \(\mu_0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} i & =\sqrt{2} A \\ B_{n e t} & =\sqrt{B_1^2+B_2^2} \\ & =\sqrt{\left(\frac{\mu_0 i}{2 R}\right)^2+\left(\frac{\mu_0 i}{2 R}\right)^2}=\sqrt{2\left(\frac{\mu_0 \mathrm{i}}{2 R}\right)^2} \\ & =\sqrt{2 \frac{\mu_0^2 \mathrm{i}^2}{4 \mathrm{R}^2}}=\sqrt{\frac{1}{2}\left(\frac{\mu_0 \mathrm{i}}{R}\right)^2} \\ & =\frac{1}{\sqrt{2}} \frac{\mu_0 \mathrm{i}}{\mathrm{R}}=\frac{1}{\sqrt{2}} \times \frac{\mu_0 \sqrt{2}}{1}=\mu_0\end{aligned}\)
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