need help with all parts A conducting rod is pulled horizontally with constant f
ID: 1581424 • Letter: N
Question
need help with all parts
A conducting rod is pulled horizontally with constant force F= 4.80 N along a set of rails separated by d= 0.320 m. A uniform magnet 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. Submit Answer Tries 0/99 The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) Submit Answer Tries 0/99 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 0/99 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 energy dissipation (power dissipated)? Submit Answer Tries 0/99Explanation / Answer
d= 0.32 m
v= 5.10 m/s
B= 0.5
=> Rate of change of area = d*v = 1.632 m^2/s
=> Rate of change of magnetic flux = B* (d*v) = 0.816 Weber/s
(a)
Since Flux is increasing (into the page)
Lenz' Law It tries to oppose the change so, using right hand thumb rule current must be produced in such a way to that B must be (from the page) induced current and hence EMF is in anti-clock wise direction
=> its +0.816 V
(b)
Now since Force * Velocity = Power
We have P = 4.8 * 5.1
But P also equals V * I
=> 4.8 * 5.1 = 0.816 * I
=> I = 30 Amps
(c)
Now R = V/I
=> R = 0.816/30
=> R = 0.0272 Ohms
(d)
Power is same again
=> P= 4.8 * 5.1
=> P = 24.48 Watts
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