Question

51. The force per meter between the two wires of a jumper cable being used to start a stalled car is 0.225 N/m. (a) What is the current in the wires, given they are separated by 2.00 cm

Answer 1

The current in both the wires is 150 A.

Step-by-step explanation:

It is given that,

Force per unit length between the two wires is given,
`(F)/(l)=0.225\ N/m`

Let I is the current in the wire when they are separated by 2 cm or 0.02 m

Force per unit length is given by :

`(F)/(l)=(\mu_o I_1I_2)/(2\pi r)`

Let
`I_1=I_2=I`

`(F)/(l)=(\mu_o I^2)/(2\pi r)`

`I=\sqrt{(F.2\pi r)/(\mu_o l)}`

`I=\sqrt{(0.225* 2\pi * 0.02)/(4\pi * 10^(-7))}`

I = 150 A

So, the current in both the wires is 150 A. Hence, this is the required solution.