MHT CET · Physics · Magnetic Effects of Current
Two concentric circular coils \(\mathrm{A}\) and \(\mathrm{B}\) have radii \(20 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) respectively lie in the same plane. The current in coil A is \(0.5 \mathrm{~A}\) in anticlockwise direction. The current in coil B so that net field at the common centre is zero, is
- A \(0.5 \mathrm{~A}\) in anticlockwise direction
- B \(0.25 \mathrm{~A}\) in anticlockwise direction.
- C \(0.25 \mathrm{~A}\) in clockwise direction.
- D \(0.125 \mathrm{~A}\) in clockwise direction.
Answer & Solution
Correct Answer
(D) \(0.125 \mathrm{~A}\) in clockwise direction.
Step-by-step Solution
Detailed explanation
Magnetic field at the centre of a circular loop
\(
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
\)
Net Magneitc field \(B_{\text {net }}=0\)
(given)
\(\begin{aligned} & \Rightarrow B_{\text {net }}=\frac{\mu_0 \mathrm{~N}_1 \times 0.5}{2 \times(0.2)}-\frac{\mu_0 \mathrm{~N}_1 \times \mathrm{x}}{2 \times(0 \cdot 1)} \\ & \Rightarrow \frac{\mu_0 \mathrm{~N}_1 \mathrm{x}}{0.2}=\frac{\mu_0 \mathrm{~N}_1 0.5}{0.4} \\ \therefore \quad & x=\frac{0.5 \times 0.2}{0.4} \\ & =0.25 \mathrm{~A} \text { in clockwise direction }\end{aligned}\)
\(
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
\)
Net Magneitc field \(B_{\text {net }}=0\)
(given)
\(\begin{aligned} & \Rightarrow B_{\text {net }}=\frac{\mu_0 \mathrm{~N}_1 \times 0.5}{2 \times(0.2)}-\frac{\mu_0 \mathrm{~N}_1 \times \mathrm{x}}{2 \times(0 \cdot 1)} \\ & \Rightarrow \frac{\mu_0 \mathrm{~N}_1 \mathrm{x}}{0.2}=\frac{\mu_0 \mathrm{~N}_1 0.5}{0.4} \\ \therefore \quad & x=\frac{0.5 \times 0.2}{0.4} \\ & =0.25 \mathrm{~A} \text { in clockwise direction }\end{aligned}\)
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