Answer:
37.4 V
Explanation:
Given:
Length of the wire, l = 2.60m
Velocity of the wire, v = 0.360m/s
Magnetic Field, B = 50.0 µT
Angle θ = 53.0°
The expression for the magnitude of induced emf for a wire is given as;
<em>ε = Blv sinθ </em>-------------(1)
Where;
<em>ε </em>is the induced emf
<em>l</em> is the length of the wire
<em>v</em> is the velocity of the wire
<em>θ</em> is the angle between the magnetic field and the normal of the loop
Substituting the values of l = 2.60m, v = 0.360m/s, B = 50.0 µT and θ = 53.0° into equation (1)
ε = 50 × 2.60 × 0.360 × sin 53.0°
ε = 50 × 2.60 × 0.360 × 0.7986
ε = 37.376 V
ε = 37.4 V
The magnitude of the induced emf between the ends of the wire is 37.4 V