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

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

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 $$B_{C} = \dfrac {\mu_{0}}{4\pi}\dfrac {2\pi l(r)^{2}}{[r^{2} + (a/2)^{2}]^{3/2}}$$$$\therefore \dfrac {B_{D}}{B_{C}} = \dfrac {4}{(4r^{2} + a^{2})^{3/2}} \times \left [\dfrac {4r^{2} + a^{2}}{4}\right ]^{3/2}$$$$= \dfrac {4}{8} = \dfrac {1}{2}$$.Physics

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