a) For a current of 10 amp in the solenoid, determine the magnetic field intensi

Question

a) For a current of 10 amp in the solenoid,

determine the magnetic field intensity and

magnetic flux density in the center ofthe solenoid

(assume it is a "long" solenoid so that the end

effects can be ignored).

b) A small circular loop of  wire 5 mm in diameter is

inserted into the center of the same solenoid as shown

(the loop and solenoid are co-axial). With a current of

3 ARMS at 60 Hz in the wire loop, determne the

voltage induced on the solenoid. (Hint: there's an easy

. way to do this and a hard way and I'm not telling you

either!)

For a current of 10 amp in the solenoid, determine the magnetic field intensity and magnetic flux density in the center ofthe solenoid (assume it is a "long" solenoid so that the end effects can be ignored). A small circular loop of wire 5 mm in diameter is inserted into the center of the same solenoid as shown (the loop and solenoid are co-axial). With a current of 3 ARMS at 60 Hz in the wire loop, determne the voltage induced on the solenoid. (Hint: there's an easy . way to do this and a hard way and I'm not telling you either!)

Explanation / Answer

a)

magnetic field B=uo*n*i

B=uo*N/l*i

where,

no.of turns,N=2000

l=1m

i=10 amp

so,

B=4*pi*10^-7*2000/1*10

B=0.02512 Tesla ,........is answer

flux density=B*A

=0.02512*pi*r^2

=0.02512*3.14*0.01^2 (d=2 cm, r=d/2=1cm)

=7.88768*10^-6 Wb ...is answer

b)

magnetic field B due to circular loop is,

B=uo*i/2*a

flux through solenoid =B*NA

=uo*i/2*a*N*pi*r^2

mutual inductance M=flux/i

=(uo/2*a)*N*pi*r^2

=(4*3.14*10^-7/2*0.0025)*2000*3.14*0.01^2

=1.577536*10^-4 wb/amp

now,

induced emf e=Mdi/dt

=M*io*w

=1.577536*10^-4*3*2*3.14*60

=0.1783 volt .....is answer

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