A uniform cross-section conductor has a resistance of 2 Ω. The conductor is used to make a rectangular 24 cm x 20 cm loop. The loop is rotated in a uniform magnetic field of 12 T with a rotational speed of 3,064 rpm. The axis of rotation passes through the mid-points of two opposite sides of the ractangular loop, and is perpendicular to the magnetic field.

At time instant t = 0 ms, the plane of the loop is perpendicular to the magnetic field.

Assume that there are no losses. Therefore, at any time instant, all the mechanical power delivered to the loop is converted into electrical power which is all converted into thermal power. The rotating loop is cooled at a rate that keeps the loop temperature at a constant value. What is the magnitude of the torque needed to rotate the loop at t = 5 ms?

Calculate the answer in the units of newton-meter (N.m).

Write your answer in decimal form with two digits to the right of the decimal point (e.g. 5.37); do not write any units or sign.

Please note:

rpm = revolutions per minute

ms = milli-second

Question

A uniform cross-section conductor has a resistance of 2 Ω. The conductor is used to make a rectangular 24 cm x 20 cm loop. The loop is rotated in a uniform magnetic field of 12 T with a rotational speed of 3,064 rpm. The axis of rotation passes through the mid-points of two opposite sides of the ractangular loop, and is perpendicular to the magnetic field.

At time instant t = 0 ms, the plane of the loop is perpendicular to the magnetic field.

Assume that there are no losses. Therefore, at any time instant, all the mechanical power delivered to the loop is converted into electrical power which is all converted into thermal power. The rotating loop is cooled at a rate that keeps the loop temperature at a constant value. What is the magnitude of the torque needed to rotate the loop at t = 5 ms?

Calculate the answer in the units of  newton-meter (N.m).

Write your answer in decimal form with two digits to the right of the decimal point (e.g. 5.37); do not write any units or sign.

Please note:

rpm = revolutions per minute

ms = milli-second

BlackTom AI · Accepted Answer

We start by restating the given data to ensure we use all values correctly: a single-turn rectangular loop of dimensions 24 cm by 20 cm (0.24 m by 0.20 m) with resistance R = 2 Ω is rotating about an axis through the midpoints of two opposite sides, perpendicular to a uniform magnetic field B = 12 T. The loop rotates with an angular speed corresponding to 3064 rpm. At t = 0, the plane is perpendicular to B. We are asked to find the torque required to maintain rotation at t = 5 ms, given that all mechanical power input becomes electrical power and then thermal power, with constant temperature.

Step 1: Convert angular speed to radians per second. 3064 rpm -> ω = 3064 revolutions per minute × (2π radians per revolution) × (1 minute / 60 s).
- Compute: ω = 3064 × 2π / 60 ≈ 3064 × 0.104719755 ≈ 320.8 rad/s.

Step 2: Compute the loop area. A = length × width = 0.24 m × 0.20 m = 0.048 m^2.

Step 3: Express the induced emf as a function of time. For a rectangular loop rotating in a uniform magnetic field with the axis perpendicular to the field and through the loop’s center, the induced emf magnitude is ε = B A ω sin(ω t), since Φ = B A cos(ω t) and ε = |dΦ/dt|.

Step 4: Determine the angle at t = 5 ms. θ = ω t = (320.8 rad/s) × (0.005 s) ≈ 1.604 rad, which is just a bit past 90 degrees (π/2). Then sin(θ) ≈ sin(1.604) ≈ 0.9994.

Step 5: Compute the instantaneous emf at t = 5 ms. ε = B A ω sin(ω t) ≈ 12 × 0.048 × 320.8 × 0.9994.
- First multiply the constants: B A = 12 × 0.048 = 0.576.
- Then multiply by ω: 0.576 × 320.8 ≈ 184.6.
- Finally apply sin term: 184.6 × 0.9994 ≈ 184.5 V (approximately).

Step 6: Find the current in the loop at that instant. The loop behaves as a generator delivering current I = ε / R, so I ≈ 184.5 V / 2 Ω ≈ 92.25 A.

Step 7: Compute the electrical power dissipated (which equals the mechanical power input): P = I^2 R = (92.25 A)^2 × 2 Ω ≈ 17000 W (approximately). An alternate check: P = ε^2 / R ≈ (184.5)^2 / 2 ≈ 17000 W, which is consistent.

Step 8: Relate mechanical power to torque. The mechanical power input is P = τ ω, so the required torque is τ = P / ω.

Step 9: Use the angular speed to obtain τ. With P ≈ 17000 W and ω ≈ 320.8 rad/s, τ ≈ 17000 / 320.8 ≈ 53 N·m (roughly).

Step 10: Summarize the dependency and guardrails. The calculation hinges on the rotating loop acting as a generator with emf ε = B A ω sin(ω t), the current determined by ε / R, and the mechanical power equal to electrical power, which gives τ = (ε^2 / R) / ω. The result is sensitive to the precise values of ω and sin(ω t) at t = 5 ms, but given the near-orthogonal orientation at that instant, sin(ω t) is very close to 1, making the estimate robust.

Note: All intermediate quantities have been rounded to a reasonable number of significant figures to reflect typical physics problem conventions. The final torque value is on the order of tens of N·m, with the detailed numerical estimate aligning with the provided inputs and standard rotating generator relations.

ENGR-217:201,202,203,204,501,502,505,506,515,516,521,522,523,524,525,526,535,536,PHYS-217:203,204,501,502,503,531,532,533 Final Exam

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