SATHEE: Electro Magnetic Induction And Alternating Currents Question 510

Electro Magnetic Induction And Alternating Currents Question 510

Question: A conducting ring of radius r with a conducting spoke OA is in pure rolling on a horizontal surface in a region having a uniform magnetic field B as shown, v being the velocity of the centre of the ring. Then the potential difference $V_{0} - V_{A}$ is:

Options:

A) $\frac{B v r}{2}$

B) $\frac{3 B v r}{2}$

C) $\frac{B v r}{2}$

D) $\frac{- 3 B v r}{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

Considering pure rolling of OA about A: the induced emf across OA will be:
$| . \vec{e} | . = \frac{B ω {(r)}^{2}}{2}$
From lenz law, 0 will be the negative end, while A will be the positive end. Hence $V_{0} - V_{A} = - \frac{B ω r^{2}}{2}$

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Electro Magnetic Induction And Alternating Currents Question 510

Question: A conducting ring of radius r with a conducting spoke OA is in pure rolling on a horizontal surface in a region having a uniform magnetic field B as shown, v being the velocity of the centre of the ring. Then the potential difference V0−VA is:

Options:

Answer:

Solution:

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Question: A conducting ring of radius r with a conducting spoke OA is in pure rolling on a horizontal surface in a region having a uniform magnetic field B as shown, v being the velocity of the centre of the ring. Then the potential difference $V_{0} - V_{A}$ is: