MHT CET · Physics · Magnetic Effects of Current
An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current I ampere and the radius of the circular loop is r metre. Then the magnetic induction B at the centre of the circular part is:
- A zero
- B infty
- C \(\frac{\mu_{\mathrm{o}} 2 \mathrm{I}}{4 \pi \mathrm{r}}(\pi+1)\)
- D \(\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{R}}(\pi-1)\)
Answer & Solution
Correct Answer
(D) \(\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{R}}(\pi-1)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& B=\left|\frac{\mu \mathrm{OI}}{2 \pi \mathrm{R}}-\frac{\mu \mathrm{OI}}{2 \mathrm{R}}\right| \\
& \mathrm{B}=\frac{\mu \mathrm{Oi}}{2 \mathrm{R}}\left(1-\frac{1}{11}\right) \\
& =\frac{\mu o \mathrm{i}}{2 \pi \mathrm{R}}(\pi-1) \\
& =\frac{\mu \mathrm{o}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{R}}(\pi-1)
\end{aligned}\)
\(\therefore\) Option D is correct answer.
& B=\left|\frac{\mu \mathrm{OI}}{2 \pi \mathrm{R}}-\frac{\mu \mathrm{OI}}{2 \mathrm{R}}\right| \\
& \mathrm{B}=\frac{\mu \mathrm{Oi}}{2 \mathrm{R}}\left(1-\frac{1}{11}\right) \\
& =\frac{\mu o \mathrm{i}}{2 \pi \mathrm{R}}(\pi-1) \\
& =\frac{\mu \mathrm{o}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{R}}(\pi-1)
\end{aligned}\)
\(\therefore\) Option D is correct answer.
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