MEC O9

Since \(a \gg b\), we can treat the smaller coil (radius \(b\)) as a magnetic dipole placed in the magnetic field generated by the larger coil (radius \(a\)).

Step 1: Magnetic Field of the Large Coil

The magnetic field \(B\) on the axis of a current-carrying loop of radius \(a\) at a distance \(x\) is given by:

$$ B = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}} $$

Step 2: Force on the Small Coil (Dipole)

The magnetic moment of the small coil is \(m = I(\pi b^2)\). The force on a dipole in a non-uniform magnetic field is given by:

$$ F = m \left| \frac{dB}{dx} \right| $$

Let’s differentiate \(B\) with respect to \(x\):

$$ \frac{dB}{dx} = \frac{d}{dx} \left[ \frac{\mu_0 I a^2}{2} (a^2 + x^2)^{-3/2} \right] $$ $$ \frac{dB}{dx} = \frac{\mu_0 I a^2}{2} \left( -\frac{3}{2} \right) (a^2 + x^2)^{-5/2} (2x) $$ $$ \frac{dB}{dx} = – \frac{3 \mu_0 I a^2 x}{2 (a^2 + x^2)^{5/2}} $$

Step 3: Calculate Force Magnitude

Substituting \(m\) and \(dB/dx\) into the force equation:

$$ F = (I \pi b^2) \times \frac{3 \mu_0 I a^2 x}{2 (a^2 + x^2)^{5/2}} $$ $$ F = \frac{3 \pi \mu_0 I^2 a^2 b^2 x}{2 (a^2 + x^2)^{5/2}} $$

Note: The power \(5/2\) can be written as \(2.5\).