MHT CET · Physics · Magnetic Effects of Current
An arc of a circle of radius ' \(R\) ' subtends an angle \(\frac{\pi}{2}\) at the centre. It carries a current \(I\). The magnetic field at the centre will be ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 I}{2 R}\)
- B \(\frac{\mu_0 I}{8 R}\)
- C \(\frac{\mu_0 I}{4 R}\)
- D \(\frac{2 \mu_0 I}{5 R}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu_0 I}{8 R}\)
Step-by-step Solution
Detailed explanation
The magnetic field due to current carrying circular arc is \(B=\frac{\mu_0 I}{2 r}\left(\frac{\theta}{2 \pi}\right)\)
Here, \(\theta=\frac{\pi}{2}\)
\(\begin{aligned}
\therefore \quad B & =\frac{\mu_0 I}{2 r}\left(\frac{1}{2 \pi} \times \frac{\pi}{2}\right) \\
B & =\frac{\mu_0 I}{2 r}\left(\frac{1}{4}\right) \\
B & =\frac{\mu_0 I}{8 r}
\end{aligned}\)
Here, \(\theta=\frac{\pi}{2}\)
\(\begin{aligned}
\therefore \quad B & =\frac{\mu_0 I}{2 r}\left(\frac{1}{2 \pi} \times \frac{\pi}{2}\right) \\
B & =\frac{\mu_0 I}{2 r}\left(\frac{1}{4}\right) \\
B & =\frac{\mu_0 I}{8 r}
\end{aligned}\)
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