Sol:

Given:

N = 500

R = \( 10\Omega \)

B = 0.14 Wb \( m^{-2} \) \( \omega = 50 rad s^{-1} \) \( A = 50 cm^{2} = 0.005 m^{2} \)

The Peak \( EMF = E_{0} = NAB\omega \)

Therefore, the peal current = \( \frac{E_{0}}{R} \) \( \Rightarrow \frac{NAB\omega}{R} \) \( \Rightarrow \frac{500 * 0.005 * 0.14 * 50}{10} \) \( \Rightarrow 1.75 A \)

Therefore the peak current through the resistance is 1.75 A

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question