Question

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

• A. 6.6 x 10-9 Weber
• B. 9.1 x 10-11 Weber
• C. 6 x 10-11 Weber
• D. 3.3 x 10-11 Weber

Answer 1

Answer: (B)

9.1 x 10-11 Weber

Answer 2

Let the current in the bigger loop be i2 and the smaller loop be i1

• ϕ1 is the flux due to the smaller loop at the bigger loop
• ϕ2 be flux due to a bigger loop at the smaller loop

The field due to the current loop 1 at an axial point -

$B_{1}=\frac{\mu_{0}I_{1}R^{2}}{2\left(d^{2}+R^{2}\right)^{3 /2}}$
Flux linked with the smaller loop 2 due to B1 is

$\phi_{2}=B_{1}A_{2}=\frac{\mu_{0}I_{1}R^{2}}{2\left(d^{2}+R^{2}\right)^{3/ 2}}\pi r^{2}$
The coefficient of mutual inductancebetween the loops is

$M=\frac{\phi_{2}}{I_{1}}=\frac{\mu_{0}R^{2}\pi r^{2}}{2\left(d^{2}+R^{2}\right)^{3 / 2}}$
Flux linked with bigger loop 1 is

$\phi_{1}=MI_{2}=\frac{\mu_{0}R^{2}\pi r^{2}l^{2}}{2\left(d^{2}+R^{2}\right)^{3 / 2}}$

Substituting the given values, we get

$\phi_{1}=\frac{4\pi\times10^{-7}\times\left(20\times10^{-2}\right)^{2}\times\pi\times\left(0.3\times10^{-2}\right)^2\times2}{2\left[\left(15\times10^{-2}\right)^{2}+\left(20\times10^{-2}\right)^{2}\right]^{3/ 2}}$

Option B is the correct answer, ϕ1 = 9.1 x 10 -11 Weber