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GUJCET · Physics · Moving Charges and Magnetism
Two concentric rings are kept in the same plane. Number of turns in each sign is 25 . Their radii are 50 cm and 200 cm and they carry electric currents of 0.1 A and 0.2 A respectively in mutually opposite directions. The magnitude of the magnetic field produced at their centre is _________ T.
- A \(4 \mu_0\)
- B \(2 \mu_0\)
- C \(\frac{10}{4} \mu_0\)
- D \(\frac{5}{4} \mu_0\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{4} \mu_0\)
Step-by-step Solution
Detailed explanation
D
From \(\quad B _{\text {centre }}=\frac{\mu_0 N I }{2 a}\)
\(B \quad= B _1- B _2\)
\(\begin{array}{ll}\therefore B=\frac{\mu_0 N}{2}\left[\frac{I_1}{a_1}-\frac{I_2}{a_2}\right] \\ \therefore B=\frac{\mu_0 \times 25}{2}\left[\frac{0.1}{0.5}-\frac{0.2}{2}\right] \\ \therefore B=\frac{\mu_0 \times 25}{2}\left[\frac{1}{5}-\frac{1}{10}\right] \\ \therefore B=\frac{\mu_0 \times 25}{2}\left[\frac{1}{10}\right] \\ \therefore B=\frac{5}{4} \mu_0\end{array}\)
From \(\quad B _{\text {centre }}=\frac{\mu_0 N I }{2 a}\)
\(B \quad= B _1- B _2\)
\(\begin{array}{ll}\therefore B=\frac{\mu_0 N}{2}\left[\frac{I_1}{a_1}-\frac{I_2}{a_2}\right] \\ \therefore B=\frac{\mu_0 \times 25}{2}\left[\frac{0.1}{0.5}-\frac{0.2}{2}\right] \\ \therefore B=\frac{\mu_0 \times 25}{2}\left[\frac{1}{5}-\frac{1}{10}\right] \\ \therefore B=\frac{\mu_0 \times 25}{2}\left[\frac{1}{10}\right] \\ \therefore B=\frac{5}{4} \mu_0\end{array}\)
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