Electromagnetism — Question 173

Original QCE Vault practice · 7 marks

Q173 · Practice questionComplex unfamiliar7 marks

QUESTION 173 (7 marks)

A conducting rod of length

0.400\,\mathrm m

slides right at

3.00\,\mathrm{m\,s^{-1}}

on parallel rails. A uniform field

0.250\,\mathrm T

points into the page. The closed circuit has resistance

2.00\,\Omega

, and rail/rod resistance and friction are negligible. Motion is maintained at constant speed.

Horizontal rails connected by a resistor at the left and a vertical sliding rod at the right. Magnetic-field crosses fill the loop; the rod moves right. The rail separation is 0.400 m. Not to scale.

Determine the induced EMF magnitude and which end of the rod is at higher potential.

[2 marks]

Calculate the current and the external force required to maintain the motion.

[3 marks]

Show numerically that mechanical input power equals electrical heating power.

[2 marks]

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WORKED SOLUTION

Practice marking scheme

7 marks

ANSWER

(a)

0.300\,\mathrm V

; the upper end is at higher potential. (b)

I=0.150\,\mathrm A

counterclockwise;

F_{\rm ext}=0.0150\,\mathrm N

rightward. (c) Both are

0.0450\,\mathrm W

Worked solution

(a) In time

\Delta t

, the swept area is

Lv\Delta t

, so

|\varepsilon|=B\Delta A/\Delta t=BLv=.250(.400)(3.00)=.300\,\mathrm V

. The magnetic force on positive charges moving right is upward, so positive charge accumulates at the upper end.

(b)

I=\varepsilon/R=.300/2.00=.150\,\mathrm A

. Current in the rod is upward, making the circuit current counterclockwise. Its magnetic force is leftward, with magnitude

BIL=.250(.150)(.400)=.0150\,\mathrm N

. The external force must be equal and rightward.

(c)

P_{\rm mech}=F_{\rm ext}v=.0150(3.00)=.0450\,\mathrm W

P_{\rm elec}=I^2R=.150^2(2.00)=.0450\,\mathrm W

. The opposing magnetic force makes the external work account for the electrical energy.

Equivalent physically justified methods and consistent equivalent units accepted.

Displayed decimals are model answers; accept appropriate significant figures and consistent rounding from stated constants.

Part a: Calculate EMF using the changing area or motional EMF.

[1 mark]

Part a: Identify the upper end as positive using magnetic-force direction.

[1 mark]

Part b: Calculate circuit current.

[1 mark]

Part b: Determine current and magnetic-force directions.

[1 mark]

Part b: Calculate the balancing external force.

[1 mark]

Part c: Calculate mechanical input power.

[1 mark]

Part c: Calculate electrical heating power and identify equality.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Electromagnetism — Question 173

QUESTION 173 (7 marks)

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