Answer:
Magnetic force, F = 0.18 N
Explanation:
It is given that,
Current flowing in the wire, I = 0.2 A
Length of the wire, L = 3 m
Magnetic field, B = 0.3 T
It is placed perpendicular to the magnetic field. We need to find the magnitude of force on the wire. It is given by :
[tex]F=ILB\ sin\theta[/tex]
[tex]F=0.2\ A\times 3\ m\times 0.3\ T\ sin(90)[/tex]
F = 0.18 N
So, the magnitude of force on the wire is 0.18 N. Hence, this is the required solution.
