A coil of wire with a current flowing thru it becomes a magnet
Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
= 4 × 10^(-10)/2.5 × 10^2
= 1.6 × 10^(-12) volt
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Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :


F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.