Training Magnetism & Electromagnetism Practice Test — Magnetism & Electromagnetism
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Practice Test — Magnetism & Electromagnetism

25 min Magnetism & Electromagnetism

Practice Test — Magnetism & Electromagnetism

This practice test covers all the major topics from the Magnetism and Electromagnetism module: magnetic fields from wires, the Lorentz force on charges and currents, solenoids, Faraday's law, and electromagnetic waves. Work through each problem step by step, applying the formulas and techniques from the previous lessons. Use the answer key to check your work and identify areas for review.

Comprehensive Practice

1. A long straight wire carries 25 A. Find the magnetic field at distances of 2 cm, 5 cm, and 10 cm.
2. A proton moves at $5 \times 10^6$ m/s perpendicular to a 1.2 T field. Find (a) the force, (b) the cyclotron radius, and (c) the cyclotron frequency.
3. A 40-cm wire carries 15 A at 45° to a 0.8 T field. Find the force.
4. A solenoid has 3000 turns, is 60 cm long, and carries 8 A. Find the field inside.
5. A 200-turn coil with area 0.05 m² rotates at 50 Hz in a 0.6 T field. Find the peak EMF.
6. A magnetic field through a 100-turn coil of area 0.01 m² drops from 0.5 T to 0 in 0.1 s. Find the average EMF.
7. Light has wavelength 600 nm. Find (a) the frequency and (b) the photon energy ($E = hf$, $h = 6.63 \times 10^{-34}$ J·s).
8. Two wires 15 cm apart carry 30 A and 10 A in the same direction. (a) Force per meter? (b) Attract or repel?
9. An EM wave has intensity $I = 500$ W/m². Find $E_0$ and $B_0$.
10. An electron ($m = 9.11 \times 10^{-31}$ kg) orbits in a 0.1 T field with a radius of 1 cm. Find its speed.
11. A toroid with 1000 turns, mean radius 20 cm, carries 4 A. Find $B$ at the mean radius.
12. A generator produces peak EMF of 120 V with 50-turn coil in a 0.5 T field at 60 Hz. What is the coil area?
Show Answer Key

1. $B(2\text{cm}) = \mu_0(25)/(2\pi\times0.02) = 250$ μT; $B(5\text{cm}) = 100$ μT; $B(10\text{cm}) = 50$ μT

2. (a) $F = (1.6\times10^{-19})(5\times10^6)(1.2) = 9.6\times10^{-13}$ N. (b) $r = (1.67\times10^{-27})(5\times10^6)/[(1.6\times10^{-19})(1.2)] = 0.0435$ m $= 4.35$ cm. (c) $f = qB/(2\pi m) = 18.3$ MHz

3. $F = (0.8)(15)(0.40)\sin45° = 3.39$ N

4. $n = 3000/0.60 = 5000$ turns/m. $B = (4\pi\times10^{-7})(5000)(8) = 50.3$ mT

5. $\mathcal{E}_{\max} = NAB\omega = (200)(0.05)(0.6)(2\pi\times50) = 1{,}885$ V

6. $|\mathcal{E}| = N|\Delta\Phi/\Delta t| = 100(0.5\times0.01)/0.1 = 5.0$ V

7. (a) $f = 3\times10^8/(600\times10^{-9}) = 5.0\times10^{14}$ Hz. (b) $E = (6.63\times10^{-34})(5.0\times10^{14}) = 3.32\times10^{-19}$ J $= 2.07$ eV

8. (a) $F/L = \mu_0(30)(10)/(2\pi\times0.15) = 4.0\times10^{-4}$ N/m. (b) Same direction → attract.

9. $E_0 = \sqrt{2\mu_0 c I} = \sqrt{2(4\pi\times10^{-7})(3\times10^8)(500)} = 614$ V/m. $B_0 = E_0/c = 2.05$ μT

10. $v = qBr/m = (1.6\times10^{-19})(0.1)(0.01)/(9.11\times10^{-31}) = 1.76\times10^8$ m/s

11. $B = \mu_0(1000)(4)/(2\pi\times0.20) = 4.0$ mT

12. $A = \mathcal{E}_{\max}/(NB\omega) = 120/[50(0.5)(2\pi\times60)] = 0.01273$ m²

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