11). Part 2 of 2: Find the magnitude of the average induced emf in the coil. Ans

Question

11). Part 2 of 2:

Find the magnitude of the average induced
emf in the coil.
Answer in units of

A long, straight wire lies on a horizontal table and carries a current of 1.33 mu A. In a vacuum, a proton moves parallel to the wire (opposite the current) with a constant velocity of 33900 m/s at a constant distance d above the wire. Find the value of d. The acceleration of gravity is 9.8 m/s2 and the permeability of free space is 1.25664 Times 10-6 T middot m/A. Ignore the magnetic field due to the Earth. What current is required in the windings of a long solenoid that has 1460 turns uniformly distributed over a length of 0.088 m in order to produce a magnetic field of magnitude 0.000364 T at the center of the solenoid? The permeability of free space is 4tt Times 10-7 T middot m/A.A 357-turn solenoid with a length of 11 cm and radius of 0.9 cm carries a current of 2 A. A second coil of 4 turns is wrapped tightly about this solenoid so that it can be considered to have the same radius as the solenoid. Find the change in the magnetic flux through the coil when the current in the solenoid increases to 4.9 A in a period of 1.32 s. The permeability of a vacuum is 4 pi Times 10-7 T .m/A transformer has input voltage and current of 11 V and 6 A respectively, and an output current of 0.8 A. If there are 1215 turns turns on the secondary side of the transformer, how many turns are on the primary side?

Explanation / Answer

1.

mg=qVB

B=mg/qv =(1.67*10^-27)*9.8/(33900)*(1.6*10^-19)

B=3.02*10^-12 T

Magnertic field due to a wire is

B=uo*I/2pid

=>d=uo*I/2pi*B =(4pi*10^-7)*(1.33*10^-6)/2pi*(3.02*10^-12)

d=0.08816 m

d=8.816 cm

2.

Magentic field of a solenoid

B=uo*N*I/L

0.000364=(4pi*10^-7)*1460*I/0.088

I=17.46 mA

4.

Transformer turns ratio

Ns/Np=Ip/Is

1215/Np =6/0.8

Np=162 turns

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