Question

Using ampere circular law, find magnetic field intensity on the axis of toroid.​

Answer 1

Answer:

Magnetic field due to a toroidal solenoid: A long solenoid shaped in the form of closed ring is called a toroidal solenoid ( or endless solemoid).

Let n be the number of turns per unit length of toroid and I the current flowing through it. The current causes the magnetic field inside the turns of the solenoid. The magnetic lines of force inside the toroid are in the form of concentric circles. By symmetry the magnetic field has the same magnitude at each point of circle and is along the tangent at every point on the circle.

(I) For points inside the core of troroid

Consider a circle of radius r in the region enclosed by turns of toroid. Now we apply Ampere's circuital law to this circular path ie.

∮B.dl=μ0I ... (i)

∮B.dl=∮Bdlcos0=B.2πr

Length of toroid =2πr

N= Number of turns in toroid =n(2πr) and current in one-turn =I

∴ Current enclosed by circular pth =(n2πr).I

∴ Equation (i) gives 

B2πr=μ0(n2πrl)⇒B=μ0I

(II) For points in the open space inside the toroid: No current flows through the Amperian loop. so I=0

∮B.dl=μ0I=0⇒∣B∣inside=0

(iii) For points in the open space exterior to the toroid: The net current entering the plane of the toroid is exactly cancelled by the net current leaving the plane of the toroid.

∮B.dl=0⇒∣B∣exterior=0

For observer, current is flowing in clockwise direction hence we will see magnetic field lines going towards South Pole.

The solenoid can be regarded as a combination of circular loops placed side by side, each behaving like a magnet moment IA, where I is the current and A area of the loop.

These magnets neutralise each other except at the ends where south and north poles appear.

Magnetic moment of bar magnet =NIA

Explanation:

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