(b) The current is anticlockwise in the loop as it enters:
E2. A connecting rod AB of mass m slides without friction over two long conducting rails separated by a distance l.
Initially, the rod is moving with a velocity v0 to the right, find:
(a) The distance covered by the rod until it comes to rest.
(b) The amount of heat generated in the resistance r during the process ?
Ans: (a) Et =Blvt
F (retarding) = ItlB =
(b) Heat generated in resistance R = Loss in Kinetic Energy
= mv02 - m(o)
E3. A small square loop of wire of side l is placed inside a large square loop of wire of side (L >> l). The loops are …. and their centres coincide. What is the mutual inductance of the system
Consider the square loop th be made up of four rods each of length L, the field at the centre, that is, at a distance from each rod will be
B = 4 x
or, B = = 2 sin450
So, the flux linked with smaller loop:
2 = B1S2 = l2I and hence,
E4. What inductance would be needed to store 1 kwh of energy in a coil carrying a 200 A current ? (1 kwh = 3.6 x 106 J)
Ans: We have I = 200 A
U = 1 kwh = 3.6 x 106 J
Now, L = [As = LI2]
=> L = L = 180 H
E5. (a) Calculate the inductance of an air core solenoid containing 300 turns if the length of the solenoid is 25 m and its cross sectional area is 4 cm2 ?
(b) Calculate the self induced e.m.f. in the solenoid if the current through it is decreasing at the rate of 50 A/s.
Ans: (a) The inductance of solenoid is given by,
Substituting the values we have
= 1.81 x 10-4 H.
(b) As t = -
Here = - 50 A/s
=> E = - (1.81 x 10-4) x (- 50)
E = 9.05 x 10-3 mv.
E6. A square wire frame with side a and a straight conductor carrying a constant current I are located in the figure. The inductance and the resistance of the frame are equal to L and R respectively. The frame was turned through 1800 about axis OO’ which is located at a distance b from the current carrying conductor. Find the electric charge which passes through the frame.
Ans: Electric charge through a loop =
q = (f - i)
Since d = a dr
; Similarly f =
E7. A conducting square loop of side a is rotated in a uniform about P as shown in the figure. Find the induced e.m.f. between P and Q and indicate the relative polarity of points P and Q.
Ans: The rotation of the ring about point P generates an e.m.f. The ring within P and Q is equivalent to a rod of length PQ.
Now, PQ = a
As we know that the e.m.f. across a rod of length l rotating with velocity w is = Bwl2
Then e.m.f. between P and Q is given by:
t = Bw(a)2
t = Bwa2