Given that,
Current, I = 400 A (downwards)
The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT.
We need to find the force exerted by the Earth’s magnetic field on the 25 m-long current. We know that the magnetic force is given by :
[tex]F=iLB\sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=400\times 25\times 30\times 10^{-6}\\\\F=0.3\ N[/tex]
The force is acting in a plane perpendicular to the current and the magnetic field i.e out of the plane.