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A solenoid coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of1.75×103A/s. For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

Short Answer

Expert verified

(a) The average magnetic flux through each turn of the inner solenoid is5.68×10-8Wb.

(b) the mutual inductance of the two solenoids is1.18×10-5H.

(c) the emf induced in the outer solenoid by changing current in inner solenoid is-0.0207V.

Step by step solution

01

Define magnetic flux and mutual induction

Magnetic flux is defined as the number of magnetic field lines passing through a given closed surface. It provides the measurement of the total magnetic field that passes through a given surface area.

When changing current in one circuit changes the magnetic flux in other circuit, an emf is induced in the latter circuit, this phenomena is called mutual induction.The emf is directly proportional to the time rate of current in former circuit. The constant so introduced removing the proportionality is called mutual inductance.

ε2=-MdI1dtwhere ε2is emf induced in second coil, M is the mutual inductance and dI1dtis the time rate of current in first coil.

Mutual inductance is a measure of the mutual inductionbetween two magnetically linked circuits, given as the ratio of the induced emf to the rate of change of current producing it.

It is expressed as,

M=ϕB2I1N2 where ϕB2is magnetic flux linked with second coil having turns N2when currentI1is flowing in first coil.

Mutual inductance between two coils depend on area of cross-section, number of turns in each coil, Space between the two coils, Permeability of medium between the two coils and Length (in case of the solenoid).

02

calculate

A solenoid coil with N2=25turns of wire is wound tightly around another coil with N1=300turns. The inner solenoid is L=25long and has a diameter of 2r=2.00cm. At a certain time, the current in the inner solenoid is I1=0.120Aand is increasing at a rate of

dI1dt=1.75×103A/s.

.

(a) For this time, magnetic field at center of solenoid is,

B1=μoN1IL

B1=4π×10-7×300×0.1200.250B1=1.81×10-4T

So, the magnetic flux linked through each turn of inner solenoid is,

ϕB=B1×πr2ϕB=1.81×10-4×π0.0200222ϕB=5.68×10-8Wb

(b) mutual inductance of the combination is calculated as,

M=μ0N1N2πr12L

M=4π×10-7×300×25×π(0.02/2)20.25M=1.18×10-5H

(c) the emf induced in the outer solenoid by the changing current in the inner solenoid is calculated as,

ε2=-MdI1dtε2=-1.18×105×1750ε2=-0.0207V

Therefore, (a)The average magnetic flux through each turn of the inner solenoid is 5.68×10-8Wb;(b) the mutual inductance of the two solenoids is 5.68×10-8Wb;(c) the emf induced in the outer solenoid by changing current in inner solenoid is-0.0207V

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