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Question
Using Ampere's Law, derive an expression for the magnetic induction inside an ideal solenoid carrying a steady current.
Solution
- Consider an ideal solenoid as shown in the figure below.
Ampere’s law applied to a part of a long ideal solenoid - The dots (.) show that the current is coming out of the plane of the paper and the crosses (×) show that the current is going into the plane of the paper, both in the coil of square cross-section wire.
- For the application of Ampere's law, an Amperian loop is drawn as shown in the figure and box.
- Using Ampere’s law,
`ointvec"B".vec"d""l" = mu_0"I"`
Over the rectangular loop abcd, the above integral takes the form
`int_"a"^"b" vec"B".vec"d""l" + int_"b"^"c" vec"B".vec"d""l" + int_"c"^"d" vec"B".vec"d""l" + int_"d"^"a" vec"B".vec"d""l" = mu_0"I"`
where, I is the net current encircled by the loop.
∴ BL + 0 + 0 + 0 = `mu_0"I"` ….(1)
Here, the second and fourth integrals are zero because `vec"B"` and `vec"d""l"` are perpendicular to each other. The third integral is zero because outside the solenoid, B = 0. - If the number of turns is n per unit length of the solenoid and the current flowing through the wire is i, then the net current coming out of the plane of the paper is
I = nLi
∴ Using equation (1)
BL = `mu_0"nLi"`
∴ B = `mu_0"ni"` ….(2)
Equation (2) is the required expression.
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