JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 13)
The magnetic moments associated with two closely wound circular coils $$\mathrm{A}$$ and $$\mathrm{B}$$ of radius $$\mathrm{r}_{\mathrm{A}}=10$$ $$\mathrm{cm}$$ and $$\mathrm{r}_{\mathrm{B}}=20 \mathrm{~cm}$$ respectively are equal if : (Where $$\mathrm{N}_{\mathrm{A}}, \mathrm{I}_{\mathrm{A}}$$ and $$\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}$$ are number of turn and current of $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively)
$$4 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$$
$$2 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$$
$$\mathrm{N}_{\mathrm{A}}=2 \mathrm{~N}_{\mathrm{B}}$$
$$\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}$$
Explanation
$$M_A=M_B$$
$${I_A}{N_A}\left( {\pi r_A^2} \right) = {I_B}{N_B}\left( {\pi r_B^2} \right)$$
$${I_A}{N_A} = 4{I_B}{N_B}$$
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