( 5 points) What is the magnetic flux through the small loop when the current th

Question

(5 points) What is the magnetic flux through the small loop when the current through the solenoid is 5.00 A?

(5 points) Write an equation for the mutual inductance of this solenoid-loop combination in terms of all or some of the following variables: L (the length of the solenoid), R (the radius of the solenoid), N (the number of turns in the solenoid), r (the radius of the small loop), I (the current in the solenoid), and various constants.

(5 points) Using the values given in the problem and your expression from 2b, determine the induced EMF in the small loop.

1) The figure to the right shows a conducting rod of length 1.75 m that is attached to two conducting wheels. The rod-wheels contraption sits on two parallel and horizontal conducting wires that are connected by a resistor (with resistance R = 0.600 ) between points a

and b so that a complete loop is formed. For the entirety of this problem, neglect any frictional effects and assume that the conducting wires and rods have negligible resistance.

(10 points) If this loop is completely immersed in a uniform magnetic field B = 0.900 T that points directly down, how much current is induced in the resistor if the rod-wheels combination moves with a constant speed of 2.85 m/s towards the resistor? Does the induced current move from point a to b or from point b to a? Explain.

(5 points) In order to keep the rod-wheels combination moving with a constant speed, how much force (magnitude and direction) must be applied to it?

(5 points) If the rod-wheels combination continues to move with its constant speed, describe what happens to the current when it moves past the resistor.

(5 points) If the rod-wheels combination stopped moving when it was 5.00 m to the left of the resistor, how much would the magnitude of the magnetic field have to increase in 5.00 sec to induce the same amount of current that was induced while the combination was moving with its constant velocity?

Explanation / Answer

1. field inside the solenoid,

B = u0 n I = u0 ( N / L ) I

flux through small loop = pi r^2 B

= pi (0.01^2) (4pi x 10^-7 x 2.50 x 10^6 x 5 / 0.5 )

= 9.87 x 10^-3 Wb

2. induced emf in loop = d ( pi r^2 u0 N I / L )/dt

= (pi r^2 u0 N / l ) dI/dt

e = M dI/dt

so mutual inductance, M = pi r^2 u0 N / L

3. M = pi x 0.01^2 x 4pi x 10^-7 x 2.50 x 10^6 / 0.50

M = 1.974 x 10^-3 H

e = M di/dt = (1.974 x 10^-3) (6 A / s)

e = 0.012 Volt  

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