A long vertical wire carries a steady 70 A current.As shown in the figure,a pair of horizontal rails are 0.20 m apart.A 20-Ω resistor connects points a and b,at the end of the rails.A bar is in contact with the rails,and is moved by an external force with a constant horizontal velocity of 0.90 m/s to the right,as shown.The bar and the rails have negligible resistance.At the instant that the bar is 0.20 m from the wire,what are the induced current in the resistor and its direction through the resistor? (μ₀ = 4π × 10^-7 T ∙ m/A) A)0.63 μA,from a to b B)0.63 μA,from b to a C)0.32 μA,from a to b D)0.32 μA,from b to a E)1.9 μA,from b to a

Question

Quizplus · Accepted Answer

When the bar is moved to the right, it generates a magnetic field perpendicular to the wire and the rails, which induces a current in the resistor.

Using the right-hand rule, we can determine that the magnetic field points into the page. The magnetic flux through the resistor is then given by:

Φ = B⋅A = B⋅d⋅w = (μ₀I/2πd)⋅d⋅w = μ₀Iw/2π

where d is the distance between the wire and the center of the rails, and w is the width of the rails.

The induced emf in the circuit is given by:

ε = -dΦ/dt = -(μ₀Iw/2π)⋅v/d = -μ₀Iwv/2πd

The induced current in the resistor is then:

I = ε/R = -μ₀Iwv/2πdR

Substituting the given values, we get:

I = -(4π×10^-7 T∙m/A)×(70 A)×(0.20 m)×(0.90 m/s)/(2π×0.20 m×20 Ω) = -0.63 μA

The negative sign indicates that the induced current flows in the opposite direction to the current in the wire, from a to b. Therefore, the best choice is A.

A Long Vertical Wire Carries a Steady 70 a Current