# Currents of equal magnitude passes 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 is I? - Physics

Sum

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/m, what is 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

d = 1.35 cm = 1.35 × 10-2

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.

Concept: Magnetic Field Produced by a Current in a Circular Arc of a Wire
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 10 Magnetic Fields due to Electric Current
Exercises | Q 8 | Page 249
SCERT Maharashtra Question Bank 12th Standard HSC Physics Maharashtra State Board 2021
Chapter 10 Magnetic Effect of Electric Current
Short Answer I | Q 4