Two circular coils X and Y, having equal number of turns and carrying currents in the same sense, subtend same solid angle at point O. If the smaller coil X is midway between O and Y and if we represent the magnetic induction due to bigger coil Y at O as `B_Y` and the due to smaller coil X at O as `B_X`, then find the ratio `B_X//B_Y`.

Options

A. `(B_(Y))/(B_(X))=1`
B. `(B_(Y))/(B_(X))=2`
C. `(B_(Y))/(B_(X))=1/2`
D. `(B_(Y))/(B_(X))=1/4`

Answer

C

Explanation

As two coil subtend the same solid angle at `O`, hence area of coil, `Y=4xx` area of coil `X`
[Solid angle =`(area)/((bot distance)^(2))]`
i.e, radius of coil `Y=2xx` radius of coil `X`
`:. B_(y)=(mu_(0))/(4pi)xx(2piI(2r)^(2))/([(2r)^(2)+(d^(2))]^(3//2))`
`B_(y)=(mu_(0))/(4pi)xx(2piI(r)^(2))/([(r)^(2)+(d/2)^(2)]^(3//2))`
`:. (B_(y))/(B_(x))=4/((4r^(2)+d^(2))^(3//2))xx[(4r^(2)+d^(2))/4]^(3//2)`
`=4/((4)^(3//2))=4/8=1/2`