CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications

An Iron Ring of Relative Permeability µR Has Windings of Insulated Copper Wire of N Turns per Meter. When the Current in the Windings is I, Find the Expression for the Magnetic Field in the Ring. - CBSE (Science) Class 12 - Physics

Login
Create free account


      Forgot password?

Question

An iron ring of relative permeability µr has windings of insulated copper wire of n turns per meter. When the current in the windings is I, find the expression for the magnetic field in the ring.

Solution

An iron ring of relative permeability µr having windings of insulated copper wire of n turns per meter can be considered as a toroid. Consider a toroidal solenoid with centre O as shown.

Suppose that
r = average radius of the toroid
I = current through the solenoid
To determine the magnetic field produced at the centre along the axis of the toroid due to current I, we imagine an Amperian loop of radius r and traverse it in the clockwise direction.
According to Ampere's circuital law, we have:

`ointvecB.vec(dl) = mu_rI`

The total current flowing through the toroid is NI, where N is the total number of turns.

`ointvecB.vec(dl) = mu_r(NI)` ......(1)

Now `vecB` and `vec(dl)` are in the same direction

`oint vecB.vec(dl) = B.ointdl`

`=> ointvecb.vec(dl) = B(2pir)` ....(2)

Comparing equations (1) and (2), we get:

`B(2pir) = mu_rNI`

`=> B = (mu_rNI)/(2pir)`

Now it is given that n is the number of turns per unit length, then

`n = N/(2pir)`

∴`B = mu_0nI` which is the expression for the magnetic field in the ring.

  Is there an error in this question or solution?
Solution An Iron Ring of Relative Permeability µR Has Windings of Insulated Copper Wire of N Turns per Meter. When the Current in the Windings is I, Find the Expression for the Magnetic Field in the Ring. Concept: Solenoid and the Toroid - the Toroid.
S
View in app×