First slide

Inductance

Question

The approximate formula expressing the inductance of two thin co-axial loops of the same radius a when their centres are separated by a distance l with l > > a is

Difficult

A

$\frac{1}{2} \frac{μ_{0} {πa}^{4}}{ℓ^{3}}$
B

$\frac{1}{2} \frac{μ_{0} a^{4}}{ℓ^{2}}$
C

$\frac{μ_{0}}{4 π} \cdot \frac{{πa}^{4}}{ℓ^{2}}$
D

$\frac{μ_{0}}{π} \cdot \frac{a^{4}}{ℓ^{3}}$

Solution

Let I = current in one loop. This magnetic flux at the centre of the other co-axial loop at a distance l from the centre of the first loop is
$B = \frac{μ_{0}}{4 π} \frac{2 I . {πa}^{2}}{{(a^{2} + ℓ^{2})}^{3 / 2}}$
The flux through the other loop is
$ϕ_{12} = B π a^{2}$
$= \frac{μ_{0}}{4 π} \cdot \frac{2 {πa}^{2} I}{{(a^{2} + ℓ^{2})}^{3 / 2}} {πa}^{2}$
$or \frac{ϕ_{12}}{I} = \frac{μ_{0}}{4 π} \cdot \frac{2 π^{2} a^{4}}{2 {(a^{2} + ℓ^{2})}^{3 / 2}}$
$= \frac{ϕ_{0} π a^{4}}{2 {(a^{2} + ℓ^{2})}^{3 / 2}}$
$∴ μ_{12} = M_{12} I$
$M_{12} = \frac{ϕ_{12}}{I} = \frac{1}{2} \frac{μ_{0} π a^{4}}{ℓ^{3}}$
$(∵ a << ℓ)$

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