Electromagnetic Induction and Alternating Current 1 Question 15

15. A rectangular frame ABCD, made of a uniform metal wire, has a straight connection between E and F made of the same wire, as shown in figure. AEFD is a square of side 1m and EB =FC=0.5m. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper and normal to it.

The rate of change of the magnetic field is 1T/s. The resistance per unit length of the wire is 1Ω/m. Find the magnitudes and directions of the currents in the segments AE, BE and EF.

(1993, 5M)

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Answer:

Correct Answer: 15. 722A(E to A),622A(B to E),122A(F to E)

Solution:

  1. Induced emf in two loops AEFD and EBCF would be

e1=dφ1dt=S1dBdt=(1×1)(1)V=1V

Similarly, e2=dφ2dt=S2dBdt=(0.5×1)(1)V=0.5V

Now, since the magnetic field is increasing, the induced current will produce the magnetic field in U direction. Hence, e1 and e2 will be applied as shown in the figure.

Kirchhoff’s first law at junction F gives

i1=i+i2

Kirchhoff’s second law in loop FEADF gives

3i1+i=1

Kirchhoff’s second law in loop FEBCF gives

2i2i=0.5

Solving Eqs. (i), (ii) and (iii), we get

i1=722A and i2=622A and i=(1/22)A

Therefore, current in segment AE is (7/22) A from E to A, current in segment BE is 6/22A from B to E and current in segment EF is (1/22) A from F to E.



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