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

Hint:

Magnetic flux is the number of magnetic field lines passing through a given closed surface. 

Answer 1

The Correct Option is B

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 inductance between 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

Answer 2

Magnetic flux is the number of magnetic field lines passing through a given closed surface. Thus magnetic flux is the product of the magnetic field and its area.

Formula used:

Flux formula, ϕ = BA

• B = Magnetic field 
• A = area.

\(B=\frac{μ_{0}IR^{2}}{2(R+x)^{\frac{3}{2}}}\) and \(A=πr^{2}\)

Complete Answer:

Circular loop radius, r = 0.3 cm = 3 x 10^-3 m

Bigger circular loop radius, R = 20 cm = 0.2 m

Distance, x = 15 cm = 0.15 m

Current flowing, I = 2.0 A

To calculate flux, we first have to calculate B,

\(B=\frac{μ_{0}IR^{2}}{2(R+x)^{\frac{3}{2}}}=\frac{4π×10^{−7}×2×0.2^{2}}{2(0.2^{2}+0.15^{2})^{\frac{3}{2}}}=3.22×10^{−6}T\)

Then,

ϕ = BA = 3.22 x 10^-6 x 3.14 x (3 x 10^-3)² = 9.1 x 10^-11 Weber

Thus option (B) is correct.

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 9.1 x 10^-11 Weber.