In the figure below, the rolling axle, 1.46 m long, is pushed along horizontal r

Question

In the figure below, the rolling axle, 1.46 m long, is pushed along horizontal rails at aconstant speed v = 10.00 m/s.A resistor R = 0.403 is connected to the rails at points a and b,directly opposite each other. (The wheels make good electricalcontact with the rails, so the axle, rails, and R form aclosed-loop circuit. The only significant resistance in the circuitis R.) A uniform magnetic field B = 0.0900 T is directed vertically downward. (a) Find the induced current I in the resistor. (inAmps)
(b) What horizontal force is required to keep the axlerolling at constant speed? (in Newtons)
In the figure below, the rolling axle, 1.46 m long, is pushed along horizontal rails at aconstant speed v = 10.00 m/s.A resistor R = 0.403 is connected to the rails at points a and b,directly opposite each other. (The wheels make good electricalcontact with the rails, so the axle, rails, and R form aclosed-loop circuit. The only significant resistance in the circuitis R.) A uniform magnetic field B = 0.0900 T is directed vertically downward. (a) Find the induced current I in the resistor. (inAmps)
(b) What horizontal force is required to keep the axlerolling at constant speed? (in Newtons)

Explanation / Answer

   a.   InducedEMF      E   =   B* v * L * sin    B   =   Strength ofmagneticfield   =   0.0900   T,   v   =   speedof conductor   =   10.00 m/s    L   =   Lengthofconductor   =   1.46   m,      =   Anglebetween B and L   =   900    E   =   0.0900* 10.00 * 1.46 * sin900   =   1.314   V    Current through theresistor   I   =   E /R   =   1.314 /0.403   =   3.260   A    b.   Force onconductor   F   =   B* I * L * sin    F   =   0.0900* 3.260 * 1.46 * sin900   =   0.428   N    E   =   0.0900* 10.00 * 1.46 * sin900   =   1.314   V    Current through theresistor   I   =   E /R   =   1.314 /0.403   =   3.260   A    b.   Force onconductor   F   =   B* I * L * sin    F   =   0.0900* 3.260 * 1.46 * sin900   =   0.428   N

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