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Question

Two circular coils $$C$$ and $$D$$ have equal number of turns and carry equal currents in the same direction in the same sense and subtend same solid angle at point $$O$$ as shown in figure. The smaller coil $$C$$ is midway between $$O$$ and $$D$$. If we represent magnetic field induction due to bigger coil and smaller coil $$C$$ as $$B_{D}$$ and $$B_{C}$$ respectively, then $$B_{D}/ B_{C}$$ is
679155_097b683d965b4bf5aa93deba480f549a.png


A
1:4
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B
1:2
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C
2:1
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D
1:1
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Solution

The correct option is B $$1 : 2$$
Since two coils subtend the same solid angle at $$O$$, hence area of coil $$D = 4x$$ area of coil $$C$$
Therefore, radius of coil $$D = 2\times$$ radius of coil $$C$$
$$\therefore B_{D} = \dfrac {\mu_{0}}{4\pi} \dfrac {2\pi /(2r)^{2}}{[(2r)^{2} + a^{2}]^{3/2}}$$
and $$B_{C} = \dfrac {\mu_{0}}{4\pi}\dfrac {2\pi l(r)^{2}}{[r^{2} + (a/2)^{2}]^{3/2}}$$
$$\therefore \dfrac {B_{D}}{B_{C}} = \dfrac {4}{(4r^{2} + a^{2})^{3/2}} \times \left [\dfrac {4r^{2} + a^{2}}{4}\right ]^{3/2}$$
$$= \dfrac {4}{8} = \dfrac {1}{2}$$.
714023_679155_ans_aac3cf81d42040ffbac100e45a08a00b.png

Physics

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