SATHEE: Electro Magnetic Induction And Alternating Currents Question 310

Electro Magnetic Induction And Alternating Currents Question 310

Question: A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ, the speed of the ring is V and the potential difference developed across the ring is [IIT JEE 1996]

Options:

A) Zero

B) $B ν π R^{2} / 2$ and M is at higher potential

C) $π R B V$ and Q is at higher potential

D) 2RBV and Q is at higher potential

Show Answer

Answer:

Correct Answer: D

Solution:

Rate of decrease of area of the semicircular ring $- \frac{d A}{d t} = (2 R) V$ According to Faraday’s law of induction induced emf $e = - \frac{d φ}{d t} = - B \frac{d A}{d t} = - B (2 R V)$ The induced current in the ring must generate magnetic field in the upward direction. Thus Q is at higher potential.

Electro Magnetic Induction And Alternating Currents Question 310

Question: A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ, the speed of the ring is V and the potential difference developed across the ring is [IIT JEE 1996]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electro Magnetic Induction And Alternating Currents Question 310

Question: A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ, the speed of the ring is V and the potential difference developed across the ring is [IIT JEE 1996]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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