4. (a) If the resistance of the resistor in Fig. 20-2 is slowly increased, what

Question

4. (a) If the resistance of the resistor in Fig. 20-2 is slowly increased, what is the direction of the current induced in the small circular loop inside the larger loop? (b) What would it be if the small loop were placed outside the larger one, to the left? 5. If the solenoid in Fig. 20-3 is being pulled away from the loop shown, in what direction is the induced current in the loop? 6. The moving rod in Fig. 20-4 is 12.0 cm long and is pulled at a speed of 15 0 cm/s. If the magnetic field is 0.800 T, calculate (a) the emf developed, and (b) the electric field felt by electrons in the rod. 7. The moving rod in Fig. 20-4 is 13.2 cm long and generates an emf of 120 mV while moving in a 0.90-T magnetic field. (a) What is its speed? (b) What is the electric field in the rod? 8.. A 500-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 10-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time?

Explanation / Answer

4)

a) counter clock wise ( According to Lenz's law)

b) There would not be any induced current.

5) counter clockwise

6)

a) emf = B*v*L

= 0.8*0.15*0.12

= 0.0144 volts or 14.4 mVolts

b) E = emf/L = 0.0144/0.12 = 0.12 N/c

7)

a) emf = B*v*L

==> v = emf/(B*L)

= 0.12/(0.9*0.132)

= 1.01 m/s or 101 cm/s

b) E = emf/L

= 0.12/0.132

= 0.909 N/c

8) induced emf = N*A*dB/dt

= N*A*(mue*N/L)*dI/dt

= (mue*N^2*A/L)*dI/dt

= (4*pi*10^-7*500^2*pi*0.0125^2/0.25)*(5-0)/0.6

= 5.135*10^-3 volts or 5.135 mVolts

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