EMI A conducting rod is pulled horizontally with constant force F-3.50 N along a

Question

EMI A conducting rod is pulled horizontally with constant force F-3.50 N along a set of rails separated by d- 0.620 m. A uniform magnetic field B 0.500 T is directed into the page. There is no friction b v: 4.30 m/s x X X x the rod and the rails, and the rod moves with constant velocity x Xxx Xx 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. The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) From your previous results, what must be the electrical resistance of the loop? (The resistance of the rails is negligible compared to the resistance of the rod, so the resistance of the loop is constant.) Submit Answer Tries o/5 The rate at which the external force does mechanical work must be equal to the rate at which energy is dissipated in the circuit. What is that rate of eneray dissipation (power dissipated)? Tries 0/5 O Type here to search /25/2018

Explanation / Answer

Given data;

Force, F = 3.50N

Length of the rod, d = 0.620m

velocity of the rod, v = 4.30m/s

F=qvB

where, q=1.6X10-19 C and B is magnetic field intensity.

B=F/qv

=3.50/(1.6X10-19X4.30)

=5.08X1018 T

To find induced emf 'E' given by Faraday, we will use

E=Bdv

=5.08X1018X0.62X4.3

=1.35X1019 V

To calculate current, we will use

F=IBd

I=F/Bd

=3.5/(5.08X1018X0.62)

=1.11X10-18 A

To calculate resistance 'R' :

R=V/I

=1.35X1019/1.11X10-18

= 1.21 X 1037 ohm

To calculate power dissipated, we will use

P=VI

=1.35X1019 X 1.11X 10-18

=14.98 Watt

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