SATHEE: Electro Magnetic Induction And Alternating Currents Question 275

Electro Magnetic Induction And Alternating Currents Question 275

Question: In a region of uniform magnetic induction $B = 10^{- 2}, t e s l a$ , a circular coil of radius 30 cm and resistance p2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is [CBSE PMT 1990]

Options:

A) 4p2 mA

B) 30 mA

C) 6 mA

D) 200 mA

Show Answer

Answer:

Correct Answer: C

Solution:

Amplitude of the current $i_{0} = \frac{e_{0}}{R} = \frac{ω N B A}{R} = \frac{2 π ν N B (π r^{2})}{R}$ $i_{0} = \frac{2 π \times 1 \times 10^{- 2} \times π {(0.3)}^{2}}{π^{2}} = 6 \times 10^{- 3} A = 6 m A$

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Electro Magnetic Induction And Alternating Currents Question 275

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