Question

Two long parallel wires separated by 5.0 mm each carry a current of 60 a. These two currents are oppositely directed. What is the magnitude of the magnetic field at a point that is between the two wires and 2.0 mm from one of the two wires?

Answer 1

To find the magnetic field at a point between two wires carrying currents in opposite directions, we calculate the separate magnetic fields and take their difference due to the opposite current direction. The result is a magnetic field magnitude of 2 x 10⁻⁵ T between the wires.

Step-by-step explanation:

To calculate the magnetic field at a point between two long, straight, parallel wires carrying currents in opposite directions, we use the formula for the magnetic field around a straight wire, given by µ0I/(2πr), where µ0 is the permeability of free space (4π x 10⁻⁷ T × m/A), I is the current, and r is the distance from the wire.

Each wire contributes to the total magnetic field at the point of interest:

1. Field from wire 1 (B₁): This wire is 2.0 mm away, so B₁ = (4π x 10⁻⁷ T × m/A)(60 A)/(2π(0.002 m)) = 6 x 10⁻⁵ T.
2. Field from wire 2 (B₂): The second wire is 3.0 mm (5.0 mm - 2.0 mm) away, so B2 = (4π x 10⁻⁷ T × m/A)(60 A)/(2π(0.003 m)) = 4 x 10⁻⁵ T.

The directions of the fields are opposite because the currents are in opposite directions. To get the net magnetic field between the wires (Bnet), we subtract the smaller field from the larger field due to the opposite direction of the currents: Bnet = B₁ - B₂ = 6 x 10⁻⁵ T - 4 x 10⁻⁵ T = 2 x 10⁻⁵ T.

The magnitude of the magnetic field at the point between the wires is 2 x 10⁻⁵ T.

Answer 2

B = 0.002 T

Step-by-step explanation:

It is given that,

Distance between two parallel wires, d = 5 mm = 0.005

Current in each wire, I = 60 A

Let B is the magnitude of the magnetic field at a point that is between the two wires and 2.0 mm from one of the two wires. Firstly, calculating the magnetic field due to first wire is given by :

`B_1=(\mu_oI)/(2\pi d)`

`B_1=(4\pi * 10^(-7)* 60)/(2\pi * 0.002)`

`B_1=0.006\ T`

The magnetic field due to second wire is given by :

`B_2=(\mu_oI)/(2\pi d)`

`B_2=(4\pi * 10^(-7)* 60)/(2\pi * 0.003)`

`B_2=0.004\ T`

Let B is the net magnetic field from wire A, so,

`B=B_1-B_2`

`B=0.006-0.004`

B = 0.002 T

Hence, this is the required solution.