A movable bar is pulled through a magnetic field with strength B = 0.2T along tw

Question

A movable bar is pulled through a magnetic field with strength B = 0.2T along two parallel rails. The field is into the page. The rails are a distance w = 4cm apart. At time t = 0 the bar is touching the rod connecting the two rails. The distance from the sliding rod to the fixed connecting bar as a function of time is given by l(t) = gamma t^4, where gamma is a constant. The velocity of the sliding rod is v(t) = dl/dt. (a) Compute the flux through the circuit formed by the bar, two rails, and the connecting rod as a function of time. Report the flux symbolically. (b) Compute the emf induced in the loop as a function of time. Report the emf symbolically. (c) At some point in the motion the sliding bar carries a current of 1.5A upward toward the top of the page. What is the magnetic force on the bar from the fixed magnetic field? Report both the magnitude and direction.

Explanation / Answer

a.) the magnetic flux is given by the dot product of the magnetic field and the area vector.

= B. A = BACos where is the angle between the area vector and the magnetic field which in our case is 0

So, = BACos 0 = BA = B l(t)w = Bt4 w = 0.2 t2 0.04 = 0.008 t4

b.) induced emf is given by the rate of change of flux

= - d /dt = - d(0.008 t4 )/dt = - 0.032 t3 the megative sign just says that the emf is induced in that direction so as to oppose the change in magnetic flux.

c.) Force on a current-carrying conductor in a magnetic field is given by

F = ILBsin where I is the current, L the length of the conductor and is the angle between the current and the magnetic field which in our case is 90 degrees.

So, F = 1.5 wBSin90 = 1.5 x 0.04 x 0.2 x 1 = 0.012 N

and the direction is given by the right hand rule. As we curl our right hand fingers from lthe direction of current to the direction of magnetic field, the thumb points towards the left.

So, the force on the rod will be acting towards the left.

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