SATHEE: Electro Magnetic Induction And Alternating Currents Question 465

Electro Magnetic Induction And Alternating Currents Question 465

Question: A sliding wire of length 0.25 m and having a resistance of $0.5 Ω$ moves along conducting guiding rails AB and CD with a uniform speed of 4 m/s. A magnetic field of 0.5 T exists normal to the plane of ABCD directed into the page. The guides are short -circuited with resistances of 4 and $2, Ω$ as shown. The current through the sliding wire is:

Options:

A) 0.27 A

B) 0.37 A

C) 1.0 A

D) 0.72A

Show Answer

Answer:

Correct Answer: A

Solution:

The induced emf across the sliding wire

$e = B v ℓ = 0.5 \times 4 \times 0.25 = 0.5, V$

The effective circuit is .

The equivalent resistance of the circuit

$r = \frac{4 \times 2}{4 + 2} + 0.5 = 1.83 Ω$

Now, $i = \frac{V}{R} = \frac{0.5}{1.83} = 0.27, A$

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Electro Magnetic Induction And Alternating Currents Question 465

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