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Question

ABC is a triangular frame made out of a metallic wire. BD is a median of the triangular frame joining the vertex B to the side AC, (Also made of the same metallic wire). It is known that AB=6 cm, BC=8 cm and AC=10 cm. The cross-sectional area of the wire is 1 mm2 and its resistivity is 24 nΩm. The triangle ABC lies in a cylindrical region of magnetic field such that the intersection of the surface of the cylinder with the plane containing the frame forms the circumcircle of triangle ABC. The magnetic field in the region varies at the rate of 0.263 T/s. What is the magnitude of the induced current in mA in the median BD of the frame?


A
5 mA
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B
10 mA
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C
15 mA
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D
20 mA
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Solution

The correct option is B 10 mA


From the above geometry we can find that,

BD=5 cm

Since, BD is the median of triangle ABC

Area(ABD)=Area(BDC)=A(let)

So, magnitude of induced emf in the loop ABD or BDC will be

E=dϕdt=AdBdt

E=(14×8×6×104)×0.263=12×0.263×104

It means we can assume that both parts of triangle is connected to a battery of emf E.
Let R be the resistance per cm length of the wire.


Now applying KVL to loops ABD and BDC

5RI15R(I1I2)6RI1+E=0

16I15I2=ER .........(1)

Similarly,

13RI25R(I1I2)=E

18I25I1=ER ..........(2)

Solving this we get

I1=23263ER, I2=21263ER

As we know R=ρla

Here, l=1 cm; a=1 mm2

R=(24×109)(1×102)1×106=24×105 Ω

I2I1=2263ER

=2263×12×0.236×10424×105

I2I1=0.01 A=10 mA

Hence, option (B) is correct.

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