A rectangular frame ABCD made of a uniform metal wire has a straight connection between E & F made of the same wire as shown in the figure. AEFD is a square of side 1 m & EB = FC = 0 . 5 m . The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper. The rate of change of the magnetic field is 1 T s - 1 , the resistance per unit length of the wire is 1 Ω m - 1 . If the currents in the segments AE and EF are I AE and I EF respectively, then what is the value of I AE I EF ?

Question

A rectangular frame ABCD made of a uniform metal wire has a straight connection between E & F made of the same wire as shown in the figure. AEFD is a square of side 1 m & EB = FC = 0 . 5 m . The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper. The rate of change of the magnetic field is 1 T s - 1 , the resistance per unit length of the wire is 1 Ω m - 1 . If the currents in the segments AE and EF are I AE and I EF respectively, then what is the value of I AE I EF ?,

MARKS App · Accepted Answer

e 1 = A d B d t = 1 × 1 × 1 = 1 V e 2 = 0 . 5 V From loop AEFDA ​ 1 I 1 - I 2 - 1 + 3 I 1 = 0 ⇒ 4 I 1 - I 2 = 1 ...(1) From loop FCBEF 0.5 - 2 I 2 + I 1 - I 2 = 0 ⇒ - I 1 + 3 I 2 = 0.5 ...(2) From (1) and (2) I 1 = 7 22 A, I 2 = 3 11 A, I 1 - I 2 = 1 22 A I AE = 7 22 A , I BE = 3 11 A , I EF = 1 22 A I AE I EF = 7

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