JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 15)
Two concentric circular loops of radii $$r_{1}=30 \mathrm{~cm}$$ and $$r_{2}=50 \mathrm{~cm}$$ are placed in $$\mathrm{X}-\mathrm{Y}$$ plane as shown in the figure. A current $$I=7 \mathrm{~A}$$ is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately :
_26th_July_Evening_Shift_en_15_1.png)
$$\frac{7}{2} \hat{k} \,\mathrm{Am}^{2}$$
$$-\frac{7}{2} \hat{k} \,\mathrm{Am}^{2}$$
$${7}\, \hat{k} \,\mathrm{Am}^{2}$$
$${-7}\, \hat{k} \,\mathrm{Am}^{2}$$
Explanation
$${\mu _1} = \pi r_1^2 \times {I_1}$$
$${\mu _2} = \pi r_2^2 \times {I_2}$$
$$\therefore$$ $${\mu _{net}} = ({\mu _2} - {\mu _1})\left( { - \widehat k} \right)$$
$$ = \pi \left( {r_2^2 - r_1^2} \right)I\left( { - \widehat k} \right)$$
$$ = 3.142 \times \left( {{{0.5}^2} - {{0.3}^2}} \right) \times 7\left( { - \widehat k} \right)$$
$$ = - {7 \over 2}\widehat k$$ Am2
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