SATHEE: Electro Magnetic Induction And Alternating Currents Question 430

Electro Magnetic Induction And Alternating Currents Question 430

Question: A player with 3 m long iron rod runs towards east with a speed of 30 km/hr. Horizontal component of earth’s magnetic field is $4 \times 10^{- 5} W b / m^{2}$ . If he is running with rod in horizontal and vertical positions, then the potential difference induced between the two ends of the rod in two cases will be [MP PET 1993]

Options:

A) Zero in vertical position and $1 \times 10^{- 3} V$ in horizontal position

B) $1 \times 10^{- 3} V$ in vertical position and zero is horizontal position

C) Zero in both cases

D) $1 \times 10^{- 3} V$ in both cases

Show Answer

Answer:

Correct Answer: B

Solution:

If player is running with rod in vertical position towards east, then rod cuts the magnetic field of earth perpendicularly (magnetic field of earth is south to north). Hence Maximum emf induced is $e = B v l = 4 \times 10^{- 5} \times \frac{30 \times 1000}{3600} \times 3 = 1 \times 10^{- 3} v o l t$ When he is running with rod in horizontal position, no field is cut by the rod, so e = 0.

Electro Magnetic Induction And Alternating Currents Question 430

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Electro Magnetic Induction And Alternating Currents Question 430

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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