dap... Problem 3. A toroidal solenoid of diameter D and of comparatively thin cr

Question

dap... Problem 3. A toroidal solenoid of diameter D and of comparatively thin cross-sectional diameter d is connected to a battery with emf 5. There are N turns of wire in solenoid, and the total resistance of the coiled wire is R. Suppose that the solenoid has been hooked up to the battery for a long time so that its current has reached a steady state. D t a. (3 points) What is the steady state value of the current? which directl b. (2 points) If one is looking down on the solenoid from above, then in is the magnetic field circulating? Clockwise or counterclockwis ude of t magni nere's determine an expression Law to

Explanation / Answer

part a:

when inductor is saturated, it will behave as a short circuit.

so total impedance in the circuit=resistance of the coil=R

hence current=emf/resistance=E/R

part b:

using right hand thumb rule, it can be seen that magnetic field is in clockwise direction.

part c:

ampere's law states that if B is magnetic field,

then

integration of (B/mu)*dl=total current enclosed

==>(B/mu)*2*pi*r=N*I

==>B=mu*N*I/(2*pi*r)

part d:

energy stored=0.5*L*I^2

where L is inductance of the coil

L is directly proportional to A/r where A is cross sectional area and r is radius of the toroid.

if the toroid is shortned to half, r becomes half and A remains unchanged

so L is doubled.

as resistance is directly proportional to length/area

and length becomes half so resistance will be halved.

then current will be doubled.

the energy stored will be 2*2^2=8 times

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