A conducting rod is pulled horizontally with constant force F4.80 N along a set

Question

A conducting rod is pulled horizontally with constant force F4.80 N along a set of rails separated by d0.320 m. A uniform magnetic field B= 0.500 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 5.10 m/s Using Faraday's Law, calculate the induced emf around the loop in the figure that is caused by the changing flux. Assign clockwise to be the positive direction for emf. 0.816 V When the magnetic field is uniform and normal to the plane of the loop, then the flux is of the product of the field and the area. In this problem it is the area that changes with time, not the field Submit Answer Incorrect. Tries 1/99 Previous Tries The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) 29.85 A Remember F=ma: since v is constant, a=0, so it must be true that the net force on the rod is zero. The pulin force on the rod due to the current through it and the magnetic field. Submit Answer Incorrect. Tries 1/99 Previous Tries force is compensating e

Explanation / Answer

a) Induced emf = B*v*d

= 0.5*5.1*0.32

= 0.816 V

b) induced emf = B*dA/dt

==> dA/dt = induced emf/B

= 0.816/0.5

= 1.632 m^2/s

c) Power delivered to the rod = power dissipated in the ckt

F*v = induced emf*I

==> I = F*v/(induced emf)

= 4.8*5.1/0.816

= 30 N

d) magnetic force = force exerted by the exetrenal source

= 4.80 N

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