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Question
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law
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Solution
- Consider a solenoid of length L having N turns.
- Solenoid’s diameter is much smaller compared to its length.
The magnetic field inside the solenoid: - Consider a rectangular loop abcd.
- From Ampere’s circuital law,
- Elemental length bc and da are perpendicular to the magnetic field.
`therefore int_"b"^"c" vec"B" * vec"dl" = int_"b"^"c" |vec"B"| vec"dl" cos 90^circ = 0`
`int_"d"^"a" vec"B" * vec"d1" = 0`
Amperian loop for solenoid
`oint_"C" vec"B" * vec"d1" = mu_0"I"_"enclosed"`
`oint_"C" vec"B"*vec"d1" = int_"a"^"b" vec"B" * vec"d1" + int_"b"^"c" vec"B"*vec"d1" + int_"c"^"d" vec"B" * vec"d1" + int_"d"^"a" vec"B" * vec"d1"`
Magnetic field outside the solenoid is zero
`int_"c"^"d" vec"B"*vec"dl" = 0`
For the path ab, `int_"a"^"b" vec"B"*vec"dl" = "B" int_"a"^"b" "dl" cos theta`
`= "B" int_"a"^"b" "dl"`
`= int_"a"^"b" vec"B" * vec"dl"` = BL
L - length of the solenoids
I - current passiing through the solenoid
N - Number of turns per unit length
`int_"a"^"b" vec"B"*vec"dl" = "BL" = mu_0"NI"`
B = `(mu_0 "NI")/"L"`
`"N"/"L" = "n"`
∴ `"N"/"L" = "n"`
∴ B = `mu_0 "nI"`
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