For a toroid N = 500, radius = 40 cm, and area of cross section `=10cm^(2)`. Find inductance

A toroid of 1000 turns has an average radius of 10 cm and area of cross section 15 cm^(2) . Calculate the self-inductance of the toroid.

Find the self-inductance of a toroidal solenoid of radius 20 cm, area of cross section 100 cm^(2) and having 500 turns. Assume that the field variations across the cross section of the toroid are negligible. Further, if a secondary coil of 300 turns is wound on this toroid and the current in the primary coil is increased from 2 A to 5 A in 0.05 s. Find the emf induced in the secondary coil.

A toroidal solenoid with an air core has an average radius of 15 cm , area of cross-section 12 cm^(2) and 1200 turns . Ignoring the field variation across the cross-section of the toroid the self-inductance of the toroid .

A toroidal solenoid with air core has an average radius of 15cm, area of cross section 12cm^(2) and has 2000 turns. Calculate the self- inductance of the toroid. Assume the field to be uniform across the cross-section of the toroid.

A toroidal solenoid with an air core has an average radius of 15 cm, area of cross-section 12 cm^(2) and 1200 turns. Obtain the self inductance of the toroid. Ignore field variations across the cross-section of the toroid. (b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obtain the induced e.m.f. in the second coil.

A solenoid of length 50 cm with 20 turns per cm and area of cross section 40 cm^(2) comletely surrounds another co-axial solenoid of the same length, area of cross seciton 25 cm^(2) with 25 turns per cm. Calculate the mutual inductance of the system.

A load of 100 N acts on a bar of length 10 cm and area of cross section 1 cm. Find the energy stored in it. (Y= 10^(11) N//m^(2))