Question

Self Inductance 

1. A solenoid used in magnetic resonance in wound from a niobium-titanium superconducting wire 2.0 mm separated by an insulating layer of negligible thickness. magnetic resonance imaging is 2.4 m long and 95 cm in diameter. It is inium superconducting wire 2.0 mm in diameter, with adjacent turns seperated by an insulating of negligible thickness.

What is the self-inductance of the MRI solenoid? ( Ans: 0.53 H) 

2. A current of 5.0 amp's is flowing in a 2.0-H inductor. The current is reduced steadily to zero in oms. What is the magnitude of the reverse emf in the inductor while the current is being turned off? (Ans: 10,000 V) 

Note: this is enough to produce a lethal shock. Note that this voltage is quite unrelated to the voltage of the battery or whatever else was supplying the inductor current. We could have a 6 V battery and still be electrocuted trying to open the circuit rapidly when a large inductance is present. 

3. A 2,000-turn solenoid is 65cm long and has a cross sectional area of 30 cm? What rate of change of current will produce a 600-V emf in this solenoid? (Ans: 25.9 A/ms)

Answer 1

1] self inductance = u0*N^2*A/l where N is number of turns, A is area of loop, l is length

= 4pi*10^-7*(2.4/2e-3)^2*[pi*0.95^2/4]/2.4

= 0.53 H answer

Answer 2

2) Given:

        I = 5A

        L = 2H

        dt = 1*10^-3

(dI/dt) = (5A/1*10^-3)

emf = L(dI/dt) = 2H(5A/1*10^-3) = 10000V