## A Rowland ring is formed of ferromagnetic material. It is circular in cross section, with an inner radius of 4.2 cm and an outer radius of 5

Question

A Rowland ring is formed of ferromagnetic material. It is circular in cross section, with an inner radius of 4.2 cm and an outer radius of 5.8 cm, and is wound with 330 turns of wire. (a) What current must be set up in the windings to attain a toroidal field of magnitude B0 = 0.25 mT? (b) A secondary coil wound around the toroid has 40 turns and resistance 6.0 Ω. If, for this value of B0, we have BM = 800B0, how much charge moves through the secondary coil when the current in the toroid windings is turned on?

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6 months 2021-08-09T08:04:20+00:00 1 Answers 1 views 0

a) 0.152A

b) 5.9×10^-5C

Explanation:

a) The magnetic feild in a tiroid in the absence of the ferro magnetic material is given by :

Bo = UonIp …eq1

Where Ip = the current in the absence of the material.

n = number of turns per unit length and is given by:

n = N/(2 pi ravg)

Where rag = (rin + rout)/2

So the number of turns per unit is :

n = N/pi(rin + rout)

Substituting into eq2

Bo = (UoNIp)/(pi ( rin + rout)

Solving for IP = pi(rin + rout)Bo / (UoN)

Substituting the given values

IP = 3.142(0.042 + 0.058)(0.2×10^-3)/ (4pi ×10^-7)×330)

Ip =( 6.284 × 10^-5)/(4.1474×10^-4)

Ip = 0.152A

b) The induced end in the 2nd coil is given by Faraday’s law as :

E = -Nd○/dt

The magnitude of the induced current in the secondary coil is given by:

I = /E/ /R

I = N/R dO/dt

The change is in the integration of current over time which gives :

q N O/r^2

The flux equals the magnetic feild multiplied by the area from the steroid feild

(rin + rout)/2 = 0.42

q = pi N Br^2/ R

The magnetic feild through the secondary coil equals the magnetic feild due to the steroid feild in addition to the magnetic field of the dipole of the ferro magnetic material.

B = Bo + Bm

But Bm = 800Bo

B = Bo + 800Bo

B = 801Bo

The charge therefore is :

q = 801BopiNr^2/R

Substituting the given values

q = [801 × 3.142 × 40 ×(0.2×10^-3)(0.42×10^-2)^2] / 6.0

q = (3.55×10^-4)/6

q = 5.92 ×10^-5C