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:** l_{1} = I_{2} = 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: **

I_{1} = I_{2} = 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)`

∴ I^{2} = `(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. **