MHT CET · Physics · Magnetic Effects of Current
A current 'I' flows in anticlockwise direction in a circular arc of a wire having \(\left(\frac{3}{4}\right)^{\text {th }}\) of circumference of a circle of radius R. The magnetic field ' \(B\) ' at the centre of circle is ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 \mathrm{I}}{3 \mathrm{R}}\) in upward direction
- B \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\) in downward direction
- C \(\frac{3 \mu_0 \mathrm{I}}{8 \mathrm{R}}\) in downward direction
- D \(\frac{3 \mu_0 \mathrm{I}}{8 \mathrm{R}}\) in upward direction
Answer & Solution
Correct Answer
(D) \(\frac{3 \mu_0 \mathrm{I}}{8 \mathrm{R}}\) in upward direction
Step-by-step Solution
Detailed explanation
Magnetic field due to \(\theta\) angle \(B=\frac{\theta}{2 \pi} \times \frac{\mu_0 I}{2 R}\)
\(\begin{aligned}
& \theta=\frac{3}{4} \times 2 \pi r=\frac{3}{2} \pi \\
& B=\frac{3 \pi}{4 \pi} \times \frac{\mu_0 I}{2 R}=\frac{3 \mu_0 I}{8 R}
\end{aligned}\)
\(\begin{aligned}
& \theta=\frac{3}{4} \times 2 \pi r=\frac{3}{2} \pi \\
& B=\frac{3 \pi}{4 \pi} \times \frac{\mu_0 I}{2 R}=\frac{3 \mu_0 I}{8 R}
\end{aligned}\)
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