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4 years ago

9

A 750-turn solenoid, 24 cm long, has a diameter of 2.3 cm . A 19-turn coil is wound tightly around the center of the solenoid. P

art A If the current in the solenoid increases uniformly from 0 to 5.1 A in 0.70 s , what will be the induced emf in the short coil during this time? Express your answer to two significant figures and include the appropriate units. |E| = nothing nothing Request Answer Provide Feedback

1 answer:

oksian1 [2.3K]4 years ago

6 0

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

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Explanation: i hope this answer your question if this is wrong or correct please let me know.also no trying to be rude but can you sent me like a thanks?

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