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A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long and has a diameter of 2.30 cm . At a certain time, the current in the inner solenoid is 0.140 A and is increasing at a rate of 1800 A/s .
For this time, calculate;

a) the average magnetic flux through each turn of the innersolenoid;
b) the mutual inductance of the two solenoids;
c) the emf induced in the outer solenoid by the changing current inthe inner solenoid
Thanks for any help you can offer! Explanation would be wonderful!

Answer :

boffeemadrid

Answer:

[tex]0.00027646\ T[/tex]

[tex]2.33\times 10^{-5}\ H[/tex]

-0.04194 V

Explanation:

[tex]N_2[/tex] = Number of turns in outer solenoid = 330

[tex]N_1[/tex] = Number of turns in inner solenoid = 22

[tex]I_1[/tex] = Current in inner solenoid = 0.14 A

[tex]\dfrac{dI_2}{dt}[/tex] = Rate of change of current = 1800 A/s

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]

r = Radius = 0.0115 m

Magnetic field is given by

[tex]B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T[/tex]

The  average magnetic flux through each turn of the inner solenoid is [tex]0.00027646\ T[/tex]

Magnetic flux is given by

[tex]\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb[/tex]

Mutual inductance is given by

[tex]M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H[/tex]

The mutual inductance of the two solenoids is [tex]2.33\times 10^{-5}\ H[/tex]

Induced emf is given by

[tex]\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V[/tex]

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

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