Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

9321924448 / 9987178554

Mon to Sat - 10 AM to 7 PM

Question

qsnImg
Asked by rbhatt16 26th June 2018, 6:08 PM
Answered by Expert
Answer:
Let I be the current passing through the conductor that has the radius of cross section R as shown in figure.
Current density J = begin mathsize 12px style fraction numerator I over denominator pi space R squared end fraction end style
Let r be radius of half of cross section. Current i through half of cross section is given by,    begin mathsize 12px style i space equals space pi space r squared cross times J space equals space pi space r squared space cross times space fraction numerator I over denominator pi space R squared end fraction space equals space I cross times r squared over R squared space
end style...........(1)
using eqn.(1) and using Ampher law, magnetic field B is given by, B = begin mathsize 12px style fraction numerator mu subscript 0 i over denominator 2 pi space r end fraction space equals space fraction numerator begin display style mu subscript 0 end style over denominator begin display style 2 pi space r end style end fraction cross times space I space r squared over R squared space equals space fraction numerator begin display style mu subscript 0 space I end style over denominator begin display style 2 pi space R squared end style end fraction space r space end style ........(2)
magnetic flux = magnetic field × Area = begin mathsize 12px style fraction numerator mu subscript 0 space I over denominator 2 space pi space R squared space end fraction integral subscript 0 superscript R r space d r space space equals space fraction numerator mu subscript 0 space I over denominator 4 space pi space space end fraction space equals space 10 to the power of negative 7 end exponent cross times space 10 space equals space 10 to the power of negative 6 end exponent space W b end style

Answered by Expert 29th June 2018, 11:52 AM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp