Obtain an expression for magnetic induction of a toroid of N no. of turns about an axis passing through its centre and perpendicular to plane.A Carnot refrigerator operates between 150 K and 200 K, calculate its coefficient of performance.

Answer: \(B=\dfrac{\mu_0NI}{2\pi r}\); coefficient of performance \(=3\)

Magnetic induction of a toroid: A toroid is a long solenoid bent into a closed ring (doughnut shape) of \(N\) closely wound turns, each carrying current \(I\). [Figure: toroid with centre O; three dashed circular Amperian loops concentric with O: loop 1 inside the inner radius, loop 2 of radius \(r\) inside the core of the windings, loop 3 outside the outer radius.]

By symmetry the field inside the windings is tangential to circles centred at O (the axis through the centre perpendicular to the plane) and has the same magnitude \(B\) at all points of a circle of radius \(r\).

Inside the core (loop 2): By Ampere’s circuital law, \(\oint\vec B\cdot d\vec l=\mu_0I_{enc}\). The loop encloses \(N\) turns, each carrying \(I\), so \(I_{enc}=NI\):

\(B\times2\pi r=\mu_0NI\Rightarrow B=\dfrac{\mu_0NI}{2\pi r}\).

If \(n=\dfrac{N}{2\pi r}\) is the number of turns per unit length, \(B=\mu_0nI\).

Open space inside (loop 1): no current is enclosed, so \(B=0\). Outside (loop 3): each turn crosses the loop twice with currents in opposite directions, so net enclosed current is zero and \(B=0\). The field is confined to the core of the toroid.

Numerical: \(T_C=150\) K, \(T_H=200\) K.

For a Carnot refrigerator, \(K=\dfrac{T_C}{T_H-T_C}=\dfrac{150}{200-150}=\dfrac{150}{50}=3\).