Question

A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer 1

Complete Question

A 590-turn solenoid is 12 cm long. The current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is
`F = 0.1602 \ N`

Step-by-step explanation:

From the question we are told that

The number of turns is
`N = 590 \ turns`

The length of the solenoid is
`L = 12 \ cm = 0.12 \ m`

The current is
`I = 36 \ A`

The diameter is
`D = 4.5 \ cm = 0.045 \ m`

The current carried by the wire is
`I = 27 \ A`

The length of the wire is
`l = 2 cm = 0.02 \ m`

Generally the magnitude of the force on this wire assuming the solenoid's field points due east is mathematically represented as

`F = B * I * l`

Here B is the magnetic field which is mathematically represented as

`B = (\mu_o * N * I )/(L)`

Here
`\mu _o` is permeability of free space with value
`\mu_ o = 4\pi *10^(-7) \ N/A^2`

substituting values

`B = (4 \pi *10^(-7) * 590 * 36 )/( 0.12)`

`B = 0.2225 \ T`

So

`F = 0.2225 * 36 * 0.02`

`F = 0.1602 \ N`