Electromagnetic Induction - 5

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

(c)

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)     E_t =Blv_t
                I_t =
F (retarding) = I_tlB =
          
          
          

(b) Heat generated in resistance R = Loss in Kinetic Energy
                                                 = mv₀² - m(o)
                                                 = mv₀²


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 sin45⁰
I
So, the flux linked with smaller loop:
₂ = B₁S₂ = l²I 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 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,
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 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