MHT CET · Physics · Magnetic Effects of Current
The magnetic moments associated with two closely wound circular coils A and B of radius \(r_A=10 \mathrm{~cm}\) and \(r_B=20 \mathrm{~cm}\) respectively are equal if \(\left(\mathrm{N}_A, \mathrm{I}_A\right.\) and \(\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}\) are number of turns and current of A and B respectively)
- A \(2 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}\)
- B \(\mathrm{N}_{\mathrm{A}}=2 \mathrm{~N}_{\mathrm{B}}\)
- C \(N_A \mathrm{I}_A=4 N_B I_B\)
- D \(4 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}\)
Answer & Solution
Correct Answer
(C) \(N_A \mathrm{I}_A=4 N_B I_B\)
Step-by-step Solution
Detailed explanation
\(\text {We know that, } \mathrm{m}=\text {NIA } \)
\( \text {Given that, } \mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \)
\( \therefore \mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}} \mathrm{A}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \mathrm{A}_{\mathrm{B}} \)
\( \therefore\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}\left[\pi\left(\mathrm{R}_{\mathrm{A}}\right)^2\right]=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}\left[\pi\left(\mathrm{R}_{\mathrm{B}}\right)^2\right] \)
\( \mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}(0.1)^2=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}(0.2)^2 \)
\( \therefore \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}\)
\( \text {Given that, } \mathrm{m}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \)
\( \therefore \mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}} \mathrm{A}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \mathrm{A}_{\mathrm{B}} \)
\( \therefore\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}\left[\pi\left(\mathrm{R}_{\mathrm{A}}\right)^2\right]=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}\left[\pi\left(\mathrm{R}_{\mathrm{B}}\right)^2\right] \)
\( \mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}(0.1)^2=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}(0.2)^2 \)
\( \therefore \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}\)
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