# Electromagnetic Induction - 5

(b) The current is anticlockwise in the loop as it enters:

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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 v

_{0}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) E

_{t}=Blv

_{t}

I

_{t}=

F (retarding) = I

_{t}lB =

(b) Heat generated in resistance R = Loss in Kinetic Energy

= mv

_{0}

^{2}- m(o)

= mv

_{0}

^{2}

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

Ans:

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 sin45

^{0}

I

So, the flux linked with smaller loop:

_{2}= B

_{1}S

_{2}= l

^{2}I and hence,

M =

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 10

^{6}J)

Ans: We have I = 200 A

U = 1 kwh = 3.6 x 10

^{6}J

Now, L = [As = LI

^{2}]

=> 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 cm

^{2}?

(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,

L =

Substituting the values we have

L =

= 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 180

^{0}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 = Bwl

^{2}

Then e.m.f. between P and Q is given by:

t = Bw(a)

^{2}

t = Bwa

^{2}