SATHEE: Electro Magnetic Induction And Alternating Currents Question 306

Electro Magnetic Induction And Alternating Currents Question 306

Question: A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity $ω$ . The induced e.m.f. between the two ends is [MP PMT 1992; Orissa JEE 2003]

Options:

A) $\frac{1}{2} B ω l^{2}$

B) $\frac{3}{4} B ω l^{2}$

C) $B ω l^{2}$

D) $2 B ω l^{2}$

Show Answer

Answer:

Correct Answer: A

Solution:

If in time t. the rod turns by an angle q, the area generated by the rotation of rod will be

$= \frac{1}{2} l \times l θ$ $= \frac{1}{2} l^{2} θ$

So the flux linked with the area generated by the rotation of rod

$φ = B (\frac{1}{2} l^{2} θ) \cos 0 = \frac{1}{2} B l^{2} θ = \frac{1}{2} B l^{2} ω, t$

And so $e = \frac{d φ}{d t} = \frac{d}{d t} (\frac{1}{2} B l^{2} ω t) = \frac{1}{2} B l^{2} ω$

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Electro Magnetic Induction And Alternating Currents Question 306

Question: A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω . The induced e.m.f. between the two ends is [MP PMT 1992; Orissa JEE 2003]

Options:

Answer:

Solution:

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Question: A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity $ω$ . The induced e.m.f. between the two ends is [MP PMT 1992; Orissa JEE 2003]