Question

Two long parallel wires 0.552 meter apart are each carrying 1.75 amperes of current, as shown(a) Find the magnitude and direction of the magnetic field at point A due to the current in the top wire. (b) Find the magnitude and direction of the force per unit length this field exerts on the bottom wire.

Answer 1

,Given,

Distance between two-wire, d=0.552 m

Current through the wires, I=1.75 A

(a) The magnetic field at a point on the bottom wire due to top wire is given by,

`B_a=(\mu_0I_1)/(2\pi d)`

Where μ₀ is the permeability of the free space.

The direction is given by the right-hand thumb rule. According to this, the direction of the magnetic field produced by top wire will into the plane of the two wires.

On substituting the known values in the above equation,

`B_a=(4\pi*10^(-7)*1.75)/(2\pi*0.552)=6.34*10^(-7)\text{ T}`

The force per unit length is given by,

`F=I_2* B_a`

The direction is given by the right-hand rule. According to this the force is directed towards the top wire.

On substituting the known values in the above equation,

`F=1.75*6.34*10^(-7)=1.11*10^(-6)\text{ N}`

Therefore the magnetic field acting on the bottom wire due to the current in the top wire is 6.34×10⁻⁷ T and the magnetic force due to this field is 1.11×10⁻⁶ N