Inductance Notes/Worksheet for PH 113 (Moloney)

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• Inductance is defined as Inductance = (flux of magnetic field)/current
• Self inductance = L = (flux of magnetic field in circuit 1)/(current in circuit 1)
• Mutual inductance = M21= (flux of magnetic field in circuit 2)/(current in circuit 1)
• or by Mutual inductance = M12 = (flux of magnetic field in circuit 1)/(current in circuit 2)
• It is true that M12 = M21
• Examples
• The units of inductance are henries. Express henrys in terms of tesla. H = T-m^2/A
• A solenoid has 200 turns and is 0.50 m long. Its radius is 0.10 m, its current is 2.0 A.
• Calculate the magnetic field inside the solenoid on its axis. (The formula is B = u_o n I.)
• 1x10^-3 T-m^2
• Calculate the total magnetic flux in the solenoid, including units. 6.3 x 10^-3 T-m^2
• Calculate the inductance of this solenoid, in henries. 3.1 x 10^-3 H
• If this same solenoid were carrying a current of 3.0 A, the inductance would be
• the same
• 1.5 times greater
• 1.5 times smaller
• none of the above
• Here we have calculated the a) self-inductance b) mutual inductance (say which it is)

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• This solenoid is surrounded by a thin coil of radius 0.15 m, and 100 turns
• Calculate the magnetic flux in the thin coil, due to the current in the solenoid (2A)
• 3.14 x 10-3 T-m^2
• Calculate the inductance which exists between the solenoid and the thin coil, in henries
• 1.57 x 10^-3 H
• If this same solenoid were carrying a current of 3.0 A, the inductance would be
• the same
• 1.5 times greater
• 1.5 times smaller
• none of the above
• This is a a) self-inductance b) mutal inductance (say which it is)

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• Faraday's law says the induced emf = - time rate of change of magnetic flux
• Inductance is magnetic flux/current
• Use these two statements to show that (induced emf) = - (inductance) (time rate of change of current) [ like emf = - L dI/dt, or emf = - M dI/dt ]
• flux = LI, emf = - d/dt (LI) = -L dI/dt
• A circuit will have both inductance and resistance.
• Some circuits have negligible resistance, some have negligible inductance
• A circuit can have high inductance and low resistance, or vice versa.
• Placing iron in a solenoid greatly increases the magnetic field
• This is because the little magnets in the iron line up with the solenoid field and make it much, much bigger (remember torque = mu cross B, which lines up a dipole with an applied B field)
• When the current is off, the soft iron magnetic regions relax back to a random state
• Does the inductance increase or decrease or stay the same when iron is inserted?
• Iron increases B and the flux, so inductance increases (a lot) when iron is present
• Inductance is like inertia.
• A body with large inertia is hard to start into motion, and also hard to stop once it is in motion.
• A circuit with large inductance is slow to build up current. But once current is built up, it is hard to stop the current, or turn it off rapidly. (A spark often results when opening a switch in a circuit with large inductance; the current 'wants' to keep on flowing, and a large emf creates the spark because of the large dI/dt)

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• A circuit has a series circuit of a 6-v battery V (text uses a curly E) , a 12-ohm resistor R , and a 1.5 millihenry inductor, L.
• Write the kirchhoff's loop equation for this circuit in terms of V, R, L, I and dI/dt.
• V-IR-LdI/dt = 0
• When I is constant, what is its value, in amperes? 1/2 A
• When I is constant, what is the induced emf in (or across the terminals of) the inductor?
• zero, because dI/dt = 0
• When we start with a switch open in this circuit, so no current is flowing, then we close the switch at t=0,
• calculate the value of the current which flows at the instant after the switch has been closed. I= 0 just after switch is closed
• What is the value of the induced emf across the terminals of the inductor at the instant just after the switch has been closed? 6V = LdI/dt, since I = 0. This induced emf momentarily completely opposes the battery and stops current flow for the first instant.
• What is the value of the current a long time after the switch has been closed? I=0

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• In the series circuit given above, a steady current is flowing, and then suddenly at t=0. the battery voltage V (text uses a curly E) becomes zero. (This could be done by closing a switch which shorts out the battery.)
• Write down the Kirchhoff's loop equation again, this time with no V in it (see p. 854)
• 0 = IR + LdI/dt
• We want to solve for I(t), so arrange for all the current stuff to be on one side of the equation
• and the time stuff on the other side of the equation. ( This is 'separation of variables'. )
• dI/I = -R/L dt
• Extra credit: integrate this equation to obtain eq. 31.12
• Text equation 31.12 is for a circuit with decaying current. ( I = Io exp(-t/tau), where tau = L/R )
• Text equation 31.10 is for a circuit with increasing current ( I = If (1-exp(-t/tau) )
• The 'time constant' is tau = L/R.
• Go through the units and show that henries/ohms = seconds
• For a rough evaluation of current vs. time, note that exp(3) = 20 and exp(-3) = 0.05
• The following questions refer to a circuit with battery, resistor and inductor where the switch is open for a long time and then closed at t=0. I did not make this clear in the original worksheet.
• If we close a switch at t=0, what percentage of the final current do we have after 3 L/R ? 95%
• Use the numerical values in the preceding example (6V, 12 ohms, 1.5 mH) and the correct equation (31.10 or 31.12) to find how many seconds after closing the switch it will take to reach a current of 0.475 A. t = 375 microseconds (3 time constants)
• What is the current at the instant just after closing the switch? I=0