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. (a) What is the flux linking the bigger loop if a current of 2.0 A flows through the smaller loop? (b) Obtain the mutual inductance of the two loops.

Solution

Given here,
Radius of bigger loop, r₁= 20 cm
Radius of small circular loop, r₂ = 0.3 cm

Distance betwen the centers of smaller and bigger loop, x = 15 cm

We know from the considerations of symmetry that M₁₂ = M₂₁.

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 take 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 B₂ at loop 2 due to crrent I₁ in loop 1 is,

B_{21} = \frac{μ_{0} I_{1} r_{1}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}}

Here x is distance between the centres.

Thus,
Flux on loop 2 due to current in loop 1 is,

ϕ_{21} = B_{21} \times A_{2}

ϕ_{21} = B_{2} {πr}_{2}^{2} = \frac{π μ_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}} I_{1}

But,

ϕ_{21} = M_{21} I_{1}

∴

M_{21} = \frac{{πμ}_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}} = M_{12}

and,

ϕ_{12} = M_{12} I_{2} = \frac{{πμ}_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}} I_{2}

Therefore,

M_{12} = M_{21} = \frac{{πμ}_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}}

Numerical:

Using the given data
b) Mutual inducatance, M₁₂ = M₂₁ =

\frac{π μ_{0} r_{1}^{2} r_{2}^{2}}{2 {(x^{2} + r_{1}^{2})}^{3 / 2}}

\frac{π \times 4 π \times 10^{- 7} \times (20 \times 10^{- 2})^{2} \times (0.3 \times 10^{- 2})^{2}}{2 \times {[(15 \times 10^{- 2})^{2} + (20 \times 10^{- 2})^{2}]}^{\frac{3}{2}}}

= 4.55 x 10^–11 H

a) Flux linking with the bigger loop when I₁ is 2.0 A is given by,

ϕ_{21} = M_{21} I_{1}

= 4.55 x 10^–11 x 2
= 9.1 x 10^–11 Wb

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Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

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