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Question
Currents of equal magnitude pass through two long parallel wires having a separation of 1.35 cm. If the force per unit length on each of the wires is 4.76 x 10-2 N, what must be I ?
Solution 1
Data: l1 = I2 = l, s = 1.35 x 10-2
F = `(mu_0/(4pi)) (2"I"_1"I"_2l)/"s" = (mu_0/(4pi)) (2"I"^2l)/"s"`
∴ `"I"^2 = "F"/"I" "s"/(2(mu_0//4pi))`
`= (4.76 xx 10^-2) (1.35 xx 10^-2)/(2 xx 10^-7) = 3.213 xx 10^3`
∴ I = `sqrt(32.13 xx 10^2)` = 56.68 A
Solution 2
Given:
I1 = I2 = I, `"F"/"L"` = 4.76 × 10-2 N
d = 1.35 cm = 1.35 × 10-2 m
To find: Electric current
Formula:
`"F"/"L" = (mu_0"I"_1"I"_2)/(2pi"d")`
Calculation:
From formula,
`4.76 xx 10^-2 = (4pi xx 10^-7 xx "I" xx "I")/(2 xx pi xx 1.35 xx 10^-2)`
∴ I2 = `(4.76 xx 10^-2 xx 1.35 xx 10^-2)/(2 xx 10^-7) = 1.35 xx 2.38 xx 10^{-2-2+7}`
I = `sqrt(1.35 xx 2.38 xx 10^3)`
= `sqrt(13.5 xx 2.38) xx 10`
= `{"anti log"(1/2(log 13.5 + log 2.38))} xx 10`
= `{"anti log"(1/2(1.1303 + 0.3766))} xx 10`
= {antilog (0.7535)} × 10
= 5.669 × 10
= 56.69 A
The electric current is 56.69 A.
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