SATHEE: Electro Magnetic Induction And Alternating Currents Question 431

Electro Magnetic Induction And Alternating Currents Question 431

Question: A coil of area 80 square cm and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of 0.05 Tesla. The maximum value of the e.m.f. developed in it is [MP PMT 1994]

Options:

A) $200 π, v o l t$

B) $\frac{10 π}{3} v o l t$

C) $\frac{4 π}{3}, v o l t$

D) $\frac{2}{3}$ volt

Show Answer

Answer:

Correct Answer: C

Solution:

$e = N B A ω; ω = 2 π f = 2 π \times \frac{2000}{60}$
$∴ e = 50 \times 0.05 \times 80 \times 10^{- 4} \times 2 π \times \frac{2000}{60} = \frac{4 π}{3}$

Electro Magnetic Induction And Alternating Currents Question 431

Question: A coil of area 80 square cm and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of 0.05 Tesla. The maximum value of the e.m.f. developed in it is [MP PMT 1994]

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 431

Question: A coil of area 80 square cm and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of 0.05 Tesla. The maximum value of the e.m.f. developed in it is [MP PMT 1994]

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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