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Question
What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero.
Solution
Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound.
Figure shows a sectional view of the toroid. The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines.
By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop.
Let the magnetic field inside the toroid be B. We shall now consider the magnetic field at S. Once again we employ
Ampere’s law in the form of,
`ointvecB.dvecI = μ_0I` Or, BL = μ0NI
Where, L is the length of the loop for which B is tangential, I be the current enclosed by the loop and N be the number of turns.
We find, L = 2πr.
The current enclosed I is (for N turns of toroidal coil) N I.
B (2πr) = µ0NI , therefore, `B = (μ_0NI)/(2pir)`
Open space inside the toroid encloses no current thus, I = 0.
Hence, B = 0
Open space exterior to the toroid:
Each turn of current carrying wire is cut twice by the loop 3. Thus, the current coming out of the plane of the paper is cancelled exactly by the current going into it. Thus,
I= 0, and B = 0.
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