FAD1022 CQ5 — Electromagnetism Quiz Set

Course quiz problems covering electromagnetic induction, Faraday's Law, Lenz's Law, motional EMF, self-inductance, mutual inductance, and transformers.

[!tip] Quick Answer Key

Q Answer Trap
1 $0.225$ A, clockwise Use vertical side (75 cm), not horizontal
2 $\pi r^2 k$, clockwise B decreasing → induced field same direction
3 $12.00$ V Mutual inductance is reciprocal
4 $5.88$ Wb Use $LI$, not voltage
5 $\pi r^2 k$, counterclockwise B out & decreasing → induced field out
6 $0.225$ A, clockwise B out of page, flux increasing
7 $2.05$ J Energy formula, not power
8 $6\ \Omega$ Two-wire system, power loss $= I^2 R$

Question 1: Motional EMF — Circuit Entering B-Field

Problem Statement: A rectangular circuit is moved at a constant velocity of $3.0$ m/s into a uniform $1.25$ T magnetic field directed out of the page. The field region is considerably wider than 50.0 cm. The circuit has dimensions 50.0 cm × 75.0 cm and resistance $12.5\ \Omega$. Determine the magnitude and direction of the current induced as it enters the field.

Given:

  • $v = 3.0$ m/s
  • $B = 1.25$ T (out of page)
  • Circuit: 50.0 cm (horizontal) × 75.0 cm (vertical)
  • $R = 12.5\ \Omega$

Solution:

Step 1: Identify the correct length The EMF is induced in the side perpendicular to both velocity and B-field: $$l = 75.0\ \text{cm} = 0.75\ \text{m}$$

Step 2: Calculate motional EMF $$\mathcal{E} = Blv = (1.25)(0.75)(3.0) = 2.8125\ \text{V}$$

Step 3: Calculate current $$I = \frac{\mathcal{E}}{R} = \frac{2.8125}{12.5} = 0.225\ \text{A}$$

Step 4: Determine direction using Lenz's Law

  • B-field is out of page, flux is increasing as circuit enters
  • Induced field must oppose this → into page
  • RHR: thumb into page, fingers curl clockwise

Answer: $0.225$ A, clockwise

[!tip] Trap Do NOT use 50 cm (horizontal side). The EMF is generated in the vertical side cutting field lines.

Question 2: Induced EMF in Circular Loop (B Into Page, Decreasing)

Problem Statement: A circular conducting loop of radius $r$ is placed in a uniform magnetic field $B$ directed into the page. The magnitude of $B$ begins to decrease at a constant rate of $k$ T/s (where $k = |dB/dt|$). Determine the induced EMF and current direction.

Given:

  • $B$ directed into page (⊗)
  • $\frac{dB}{dt} = -k$ (decreasing)

Solution:

Step 1: Calculate EMF using Faraday's Law $$|\mathcal{E}| = \left|\frac{d\Phi}{dt}\right| = A\left|\frac{dB}{dt}\right| = \pi r^2 k$$

Step 2: Determine direction using Lenz's Law

  • Original B is into page and decreasing
  • Induced field must reinforce B → into page
  • RHR: thumb into page, fingers curl clockwise

Answer: EMF = $\pi r^2 k$, current flows clockwise

[!tip] Key Rule "Decrease → Same": When flux decreases, induced field is in the same direction as original B.

Question 3: Mutual Inductance — Reciprocity Principle

Problem Statement: Two concentric circular loops are shown. When current in the inner loop changes by $1.1$ A/s, an EMF of 3 V is induced in the outer loop. Calculate the EMF induced in the inner loop when the current in the outer loop changes by $4.4$ A/s.

Given:

  • $\frac{dI_1}{dt} = 1.1$ A/s → $\mathcal{E}_2 = 3$ V
  • $\frac{dI_2}{dt} = 4.4$ A/s → Find $\mathcal{E}_1$

Solution:

Step 1: Find mutual inductance from first scenario $$\mathcal{E}_2 = M\frac{dI_1}{dt} \Rightarrow 3 = M(1.1) \Rightarrow M = \frac{3}{1.1}$$

Step 2: Apply reciprocity for second scenario $$\mathcal{E}_1 = M\frac{dI_2}{dt} = \frac{3}{1.1} \times 4.4 = 3 \times 4 = 12\ \text{V}$$

Answer: $12.00$ V

[!tip] Key Principle Mutual inductance is reciprocal: $M_{12} = M_{21} = M$. The same $M$ applies in both directions.

Question 4: Magnetic Flux from Self-Inductance

Problem Statement: A single turn of a coil with self-inductance $77$ mH is connected to a power supply of $19.6$ V, resulting in a current of $76.3$ A. Calculate the magnetic flux linked with the coil.

Given:

  • $N = 1$ turn
  • $L = 77$ mH = $0.077$ H
  • $V = 19.6$ V
  • $I = 76.3$ A

Solution:

Step 1: Use self-inductance definition $$L = \frac{N\Phi}{I} \Rightarrow \Phi = \frac{LI}{N}$$

Step 2: Calculate flux (N = 1 for single turn) $$\Phi = LI = (77 \times 10^{-3})(76.3) = 5.8751 \approx 5.88\ \text{Wb}$$

[!note] The voltage and resistance ($V = IR$ → $R = 19.6/76.3 \approx 0.257\ \Omega$) are consistent but not needed for flux calculation.

Answer: $5.88$ Wb

[!tip] Trap Do NOT use $P = VI$ or try to incorporate the voltage. For steady-state DC, flux depends only on $L$ and $I$.

Question 5: Induced EMF in Circular Loop (B Out of Page, Decreasing)

Problem Statement: A circular conducting loop of radius $r$ is placed in a uniform magnetic field $B$ directed out of the page. The magnitude of $B$ begins to decrease at a constant rate of $k$ T/s. Determine the induced EMF and current direction.

Given:

  • $B$ directed out of page (⊙)
  • $\frac{dB}{dt} = -k$ (decreasing)

Solution:

Step 1: Calculate EMF $$|\mathcal{E}| = A\left|\frac{dB}{dt}\right| = \pi r^2 k$$

Step 2: Determine direction

  • Original B is out of page and decreasing
  • Induced field must reinforce B → out of page
  • RHR: thumb out of page, fingers curl counterclockwise

Answer: EMF = $\pi r^2 k$, current flows counterclockwise

[!tip] Comparison Compare with Q2: Same formula, different direction based on original B-field orientation.

Question 6: Motional EMF — Circuit Entering B-Field (Variant)

Problem Statement: Same as Question 1: rectangular circuit entering B-field out of page.

Given:

  • $v = 3.0$ m/s, $B = 1.25$ T, $l = 0.75$ m, $R = 12.5\ \Omega$

Solution: Same calculation as Q1: $$\mathcal{E} = Blv = (1.25)(0.75)(3.0) = 2.8125\ \text{V}$$ $$I = \frac{2.8125}{12.5} = 0.225\ \text{A}$$

Direction: Flux out of page is increasing → oppose with field into page → clockwise

Answer: $0.225$ A, clockwise

Question 7: Energy Stored in Inductor

Problem Statement: A single turn of a coil with self-inductance of $73$ mH is connected to a $1.8$ V battery, resulting in a current of $7.5$ A. Calculate the magnetic energy stored in the coil.

Given:

  • $L = 73$ mH = $0.073$ H
  • $V = 1.8$ V
  • $I = 7.5$ A

Solution:

Step 1: Use energy formula for inductor $$U = \frac{1}{2}LI^2$$

Step 2: Calculate $$U = \frac{1}{2}(0.073)(7.5)^2 = \frac{1}{2}(0.073)(56.25) = 2.053\ \text{J} \approx 2.05\ \text{J}$$

Answer: $2.05$ J

[!tip] Trap Do NOT use $P = VI = 1.8 \times 7.5 = 13.5$ W (power, not energy). The voltage and current are only needed to verify the circuit is consistent.

Question 8: Transmission Line Resistance (Transformer Power Loss)

Problem Statement: A power plant generates electricity stepped up by a transformer. At the end of the line, a second transformer receives the power. Given:

  • Power output from Transformer 1: $500$ kW
  • Transmission Voltage: $10$ kV
  • Power received at Transformer 2: $485$ kW
  • Two-wire transmission system

Calculate the total resistance of the transmission cable.

Given:

  • $P_{out} = 500$ kW = $500,000$ W
  • $V = 10$ kV = $10,000$ V
  • $P_{received} = 485$ kW = $485,000$ V
  • Two-wire system

Solution:

Step 1: Calculate power lost $$P_{loss} = 500,000 - 485,000 = 15,000\ \text{W}$$

Step 2: Calculate current in transmission line $$I = \frac{P}{V} = \frac{500,000}{10,000} = 50\ \text{A}$$

Step 3: Calculate total resistance $$P_{loss} = I^2 R \Rightarrow 15,000 = (50)^2 R = 2,500R$$ $$R = \frac{15,000}{2,500} = 6\ \Omega$$

Answer: $6\ \Omega$

[!tip] Trap Remember this is a two-wire system. The calculated $R = 6\ \Omega$ is the total loop resistance. Do not divide by 2.

Key Formulas Used

Electromagnetic Induction: $$\mathcal{E} = -N\frac{d\Phi_B}{dt} = -N\frac{d(BA\cos\theta)}{dt}$$ $$\Phi_B = BA\cos\theta\ \text{(magnetic flux)}$$

Motional EMF: $$\mathcal{E} = Blv$$

Self-Inductance: $$L = \frac{N\Phi_B}{I}$$ $$\mathcal{E}_L = -L\frac{dI}{dt}$$ $$U = \frac{1}{2}LI^2\ \text{(energy stored)}$$

Mutual Inductance: $$M = \frac{N_2\Phi_{21}}{I_1} = \frac{N_1\Phi_{12}}{I_2}$$ $$\mathcal{E}_2 = -M\frac{dI_1}{dt}$$

Transformers (Power Loss): $$P_{loss} = I^2R$$ $$I = \frac{P_{transmitted}}{V_{transmission}}$$

Lenz's Law Direction Rules:

Flux Change Induced Field Direction Current Direction
B increasing Opposite to B Opposite
B decreasing Same as B Same

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