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Question
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side an as shown in the figure.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.
Solution
(a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).
Magnetic flux associated with elementdy, `"d"phi` = BdA
Where,
dA = Area of element dy = a dy
B = Magnetic field at distance y
= `(mu_0"I")/(2pi"y")`
I = Current in the wire
`mu_0` = Permeability of free space = 4π × 10−7 T m A−1
∴ `"d"phi = (mu_0"Ia")/(2pi)("dy")/"y"`
`phi = (mu_0"Ia")/(2pi)int("dy")/"y"`
y tends from x to a + x
∴ `phi = (mu_0"Ia")/(2pi) int_"x"^("a" + "x")("dy")/"y"`
= `(mu_0"Ia")/(2pi) [log_"e""y"]_"x"^("a" + "x")`
= `(mu_0"Ia")/(2pi) log_"e"(("a" + "x")/"x")`
for mutual inductunce M, the flux is given as :
`phi = "MI"`
∴ MI = `(mu_0"Ia")/(2pi) log_"e"("a"/"x" + 1)`
M = `(mu_0"a")/(2pi) log_"e"("a"/"x" + 1)`
(b) Emf induced in the loop, e = B’av = `((mu_0"I")/(2pi"x"))"av"`
Given,
I = 50 A
x = 0.2 m
a = 0.1 m
v = 10 m/s
`"e" = (4pi xx 10^-7 xx 50 xx 0.1 xx 10)/(2pi xx 0.2)`
`"e" = 5 xx 10^-5` V
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