Question

$_{_{}}$ An alternating current drives a solenoid so that the magnetic field inside obeys the equation B(t)=Bmax(sin(2 $\Pi$ ft), where Bmax=.03 T and f= 50Hz. a.)How much EMF is generated in the solenoid at time t=0? b.) If positive B(t) indicates magnetic field out of the page, what is the direction of induced EMF in the solenoid at time t=0, clockwise or counterclockwise?

Answer 1

B= Bmax[sin(2πft)]

dB/dt = d/dt[Bmax[sin(2πft)] = Bmax*2π[cos (2πft)]

induced EMF in the solenoid = dφ/dt = d(BA)/dt = A*dB/dt = A* Bmax*2π [cos (2πft)] = A*0.03*2*3.14 [cos (2*3.14*50*0)] = A*0.03*2*3.14[cos(0] = A*0.03*2*3.14*1 = A*0.1884 = 0.1884A …..(A is area of the solenoid loop, which is not given in the problem)

The magnetic field is pointing out of the paper and rising.   The coil reacts to oppose this rising outward field by generating a field that points into the paper. To do this, the current must flow to the left through the resistor.