A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

Explanation

We know from the considerations of symmetry that `M_(12)= M_(21)`. Direct calculation of flux linking the bigger loop due to the field by the smaller loop will be difficult to handle. Instead, let us calculate the flux through the smaller loop due to a current in the bigger loop. The smaller loop is so small in area that one can takes the simple formula for field B on the axis of the bigger loop and multiply B by the small area of the loop to calculate flux without much error. Let 1 refer to the bigger loop and 2 the smaller loop. Field `B_(2)` at 2 due to `I_(1)` in 1 is

`B_(2)= (mu_(0)I_(1)r_(1)^(2))/(2(x^(2) + r_(1)^(2))^(3//2))`
Here x is distance between the centres
`phi_(2)= B_(2)pi r_(2)^(2)= (pi mu_(0)r_(1)^(2) r_(2)^(2))/(2(x^(2) + r_(1)^(2))^(3//2)) I_(1)`
But `phi_(2)= M_(21)I_(1)`
`therefore M_(21)= (pi mu_(0)r_(1)^(2) r_(2)^(2))/(2(x^(2) + r_(1)^(2))^(3//2)) = M_(12)`
`therefore phi_(1)= M_(12)I_(2)= (pi mu_(0) r_(1)^(2) r_(2)^(2))/(2(x^(2) + r_(1)^(2))^(3//2)) I_(2)`
Using the given data `M_(12)= M_(21)= 4.55 xx 10^(-11)H`
`phi_(1)= 9.1 xx 10^(-11)Wb`