Tutorial 11 — Self & Mutual Induction
Course: FAD1022 Basic Physics 2
Semester: 2 2025/2026
Centre: Centre for Foundation Studies in Science, Universiti Malaya (PASUM)
Question 1
An air-cored solenoid with radius of 3.0 cm and length 7.0 cm has 800 turns. If a current of 0.5 A flows through it, determine the:
a) Self-inductance of the solenoid b) Energy stored in the inductor
Question 2
The current in a coil change from 10 A to zero in 0.50 s. If the average induced emf is 320 V, determine the energy stored in an inductor when a current of 10 A exists in the coil.
Question 3
An adjacent coil with 12 turns experiences an induced emf of 6.0 mV when the current in the primary coil changes at a rate of $2.0 \text{ A/s}$.
Calculate:
a) Determine the mutual inductance of the combination b) Calculate the rate of magnetic flux produced in secondary coil
Question 4
A solenoid with a cross-sectional area of $5 \text{ cm}^2$ has 1000 turns and is 35 cm long. A second coil is wrapped around it and has 400 turns with the same length and area.
Calculate:
a) What is the mutual inductance of each inductor? b) If the current increases from 15 A to 35 A in 200 microseconds, what is the emf induced in the second coil?
Question 5
Based on information in the figure (transformer circuit), calculate the:
a) Transmission current through the transmission cable b) Input power in transformer T1 c) Final output power in transformer T2 d) Efficiency of overall system from input power transformer T1 to output power transformer T2
Additional Questions
Question A6
A typical laptop computer consumes 84 W, 12 A to operate. To draw this current from a 240V AC source, the primary coil of a transformer with 1200 turns is connected to the AC source, and the secondary coil powers the laptop.
What is the number of turns in the secondary coil?
Answer: 35 turns
Question A7
A power efficient wireless charger operates at 20 V, 60-70 Hz can be used to charge various devices. If the ratio of coil turns of wireless charger to handphone is 4:1 and the current flows in handphone is 2 A, calculate the:
a) Resistance in the handphone — ans: 2.5 Ω b) Current flow in the wireless charger — ans: 0.5 A
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Related Concepts
- Inductance & Transformers
- Self-Inductance ($L = \frac{\mu_0 N^2 A}{l}$)
- Mutual Inductance ($M = \frac{\mu_0 N_1 N_2 A}{l}$)
- Inductor
- Induced EMF in Inductor ($\varepsilon = -L\frac{dI}{dt}$)
- Energy Stored in Inductor ($U = \frac{1}{2}LI^2$)
- Transformer
- Turns Ratio ($\frac{V_s}{V_p} = \frac{N_s}{N_p}$)
- Transformer Efficiency