smc-3705-10-Generated emf

PHYSICS, M6 EQ-Bank 26

A solenoid was connected to a data logger to measure voltage. A magnet was dropped through the solenoid from above as shown.

On the axes provided, sketch a graph showing the change in voltage as the magnet falls completely through the solenoid. (3 marks)

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Points to note on graph:

→ First peak negative, longer wavelength and shorter amplitude.

→ Second peak positive, shorter wavelength and higher amplitude.

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Points to note on graph:

→ First peak negative, longer wavelength and shorter amplitude.

→ Second peak positive, shorter wavelength and higher amplitude.

PHYSICS, M6 2015 HSC 12 MC

A simple AC generator was connected to a cathode ray oscilloscope and the coil was rotated at a constant rate. The output is shown on this graph.

Which of the following graphs best represents the output if the rate of rotation is decreased to half of the original value?

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`D`

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Halving the rate of rotation of the bar magnet:

→ Doubles the period of the output graph (eliminate A and B).

→ Halves the rate of change of flux through the coil of the generator.

→ Halves of the maximum output voltage.

`=>D`

♦ Mean mark 52%.

PHYSICS, M6 2017 HSC 27

The diagram shows an electric circuit in a magnetic field directed into the page. The graph shows how the flux through the conductive loop changes over a period of 12 seconds.

Calculate the maximum magnetic field strength within the stationary loop during the 12-second interval. (2 marks)

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Calculate the maximum voltage generated in the circuit by the changing flux. In your answer, indicate the polarity of the terminals `P` and `Q` when this occurs. (3 marks)

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a. `2.1\ text{T}`

b. `0.3 text{V}`

→ Terminal `P` will be positive and terminal `Q` will be negative (Lenz’s Law).

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a.	`Phi`	`=BA`
	`B`	`=(Phi)/(A)=(Phi)/(pir^(2))`
	`B_max`	`=(0.6)/(pi xx(0.3)^(2))`
		`=2.1\ text{T}`

♦♦ Mean mark (a) 38%.

b. Voltage (emf) = time rate of flux

→ The induced emf is at a maximum when the rate of change of flux is a maximum.

→ From the graph, this occurs at t = 10 – 12 s (steepest gradient).

`epsilon`	`=-(Delta Phi)/(Delta t)`
	`=-((-0.6)/(2))`
	`=0.3\ text{V}`

→ Terminal `P` will be positive and terminal `Q` will be negative (Lenz’s Law).

♦♦ Mean mark (b) 27%.

PHYSICS, M6 2020 HSC 28

A metal rod sits on a pair of parallel metal rails, 20 cm apart, that are connected by a copper wire. The rails are at 30° to the horizontal.

The apparatus is in a uniform magnetic field of 1 T which is upward, perpendicular to the table.

A force, `F`, is applied parallel to the rails to move the rod at a constant speed along the rails. The rod is moved a distance of 30 cm in 2.5 s.

Show that the change in magnetic flux through the circuit while the rod is moving is approximately `5.2 × 10^(-2)` Wb. (2 marks)

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Calculate the emf induced between the ends of the rod while it is moving, and state the direction of flow of the current in the circuit. (2 marks)

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The experiment is repeated without the magnetic field.
Explain why the force required to move the rod is different without the magnetic field. (3 marks)

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a. `phi=5.2 xx10^(-2)` Wb

b. `epsi=2.1 xx10^(-2) V`

The direction of the induced current is anticlockwise as viewed from above.

c. Explanation:

→ When the magnetic field is present, the induced current results in a force acting on the rod which opposes its motion (Lenz’s Law).

→ Additionally, the force required to move the rod must also overcome the downwards gravitational force.

→ Without the magnetic field, there is no opposing force due to the induced current so the force applied only needs to overcome gravity.

→ Hence, the force required to move the rod is less without the magnetic field.

Show Worked Solution

a.	`phi`	`=BA cos theta`
		`=1xx(0.3 xx0.2)xx cos 30^(@)`
		`=0.05196…`
		`~~5.2 xx10^(-2)` Wb

b.	`epsi`	`=-N(Delta phi)/(t)`
		`=-1xx(5.2 xx10^(-2))/(2.5)`
		`=0.208…`
		`=2.1 xx10^(-2)` V (V`>`0)

The direction of the induced current is anticlockwise as viewed from above (Lenz’s Law).

♦ Mean mark (b) 51%.

c. Explanation:

→ When the magnetic field is present, the induced current results in a force acting on the rod which opposes its motion (Lenz’s Law).

→ Additionally, the force required to move the rod must also overcome the downwards gravitational force.

→ Without the magnetic field, there is no opposing force due to the induced current so the force applied only needs to overcome gravity.

→ Hence, the force required to move the rod is less without the magnetic field.

♦ Mean mark (c) 50%.

A stationary coil of 35 turns and cross-sectional area of 0.02 m² is placed between two electromagnets, and connected to a voltmeter as shown. The electromagnets produce a uniform magnetic field of 0.15 T through the coil.

The magnitude of the magnetic field is then reduced to zero at a constant rate over a period of 0.4 s.

Calculate the magnitude of the emf induced in the coil. (3 marks)

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0.3 V

Show Worked Solution

`Phi`	`=BA`
	`=0.15 xx0.02`
	`=0.003` Wb

`epsi`	`=-N(Delta Phi)/(Delta t)`
	`=-35((0-0.003))/(0.4)`
	`=0.3` V