MHT CET · Physics · Magnetic Effects of Current
The magnetic field at a point \(\mathrm{P}\) situated at perpendicular distance ' \(R\) ' from a long straight wire carrying a current of \(12 \mathrm{~A}\) is \(3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\). The value of ' \(R\) ' in \(\mathrm{mm}\) is \(\left[\mu_0=4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Am}\right]\)
- A \(0.08\)
- B \(0.8\)
- C \(8\)
- D \(80\)
Answer & Solution
Correct Answer
(D) \(80\)
Step-by-step Solution
Detailed explanation
Using Biot-Savart's Law,
\(\begin{aligned}
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{R}} \\
\therefore \quad \mathrm{R} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{B}} \\
& =\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}} \\
& =8 \times 10^{-2} \mathrm{~m} \\
& =80 \mathrm{~mm}
\end{aligned}\)
\(\begin{aligned}
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{R}} \\
\therefore \quad \mathrm{R} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{B}} \\
& =\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}} \\
& =8 \times 10^{-2} \mathrm{~m} \\
& =80 \mathrm{~mm}
\end{aligned}\)
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