Question

A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Answer 1

A Rowland ring is just a toroid which produces magnetic fields. For calculating the magnetic field produced by a toroid, we need permeability of the material of the core, number of turns per unit length of the toroid and the amount of current flowing through the toroid.
Formula used:
The magnetic field produced by a toroid is given as
$B = \mu nI{\text{ }}...\left( i \right)$
Here B represents the magnetic field produced by the toroid, $\mu$ is the permeability of the material of the core of the toroid, n represents the number of turns in the toroid per unit length while I is the current flowing through the coils of the toroid.

Detailed step by step solution:
We are given a Rowland ring which is a toroid whose core is made of magnetic material. We are given the radius of the ring whose value is
$r = 15cm = 0.15m$
The number of turns in the coil is given as
$N = 3500$
But we need no. of turns per unit length of the toroid which can be obtained as follows:
$n = \dfrac{N}{{2\pi r}} = \dfrac{{3500}}{{2\pi \times 0.15}}$

We are given the relative permeability of the ferromagnetic core of the toroid.
${\mu _r} = 800$
Now we can calculate the permeability of the core as follows:
$\mu = {\mu _0}{\mu _r} = 800{\mu _0}$
Here ${\mu _0} = 4\pi \times {10^{ - 7}}$
The current flowing through the core is given as
$I = 1.2A$
Now we can calculate the magnetic field produced in the toroid by using equation (i) in the following way.
$B = \mu nI = 800 \times 4\pi \times {10^{ - 7}} \times \dfrac{{3500}}{{2\pi \times 0.15}} \times 1.2 = 4.48T$
This is the required answer.

Note: A toroid is basically a solenoid which has been folded in a circular shape. The magnetic flux is produced in a toroid by the phenomenon of magnetic induction. The direction of the produced field can be determined by using the right hand rule.