Questions
A toroidal solenoid with an air core has an average radius of 15 cm, area of cross section and 1200 turns. Ignoring the field variation across the cross-section of the toroid, the self inductance of the toroid is
detailed solution
Correct option is C
L=(μ0n2)×(2πR×A) =4π×10−7×(1200)22π×0.15×12×10−4H =2×144×120.15×10−7H=2.3mHTalk to our academic expert!
Similar Questions
A coil is wound as a transformer of rectangular cross-section. If all the linear dimensions of the transformer are increased by a factor 2 and the number of turns per unit length of the coil remain the same, the self inductance increased by a factor of
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
