Question 18.QE.4: Hand-powered computer To power computers in locations not re......
Hand-powered computer
To power computers in locations not reached by any power grid, engineers have developed hand cranks that rotate a 100-turn coil in a strong magnetic field. In a quarter-turn of the crank of one of these devices, the magnetic flux through the coil’s area changes from 0 to 0.10 T ⋅ m² in 0.50 s. What is the average magnitude of the emf induced in the coil during this quarter-turn?
Represent mathematically Faraday’s law [Eq. (18.3)] can be used to determine the average magnitude of the emf produced in the coil:
\varepsilon_{\text {in }}=N\left|\frac{\Delta \Phi}{\Delta t}\right|=N\left|\frac{\Delta[B A \cos \theta]}{\Delta t}\right| (18.3)
\varepsilon_{\mathrm{in}}=N\left|\frac{\Delta \Phi}{\Delta t}\right|=N\left|\frac{\Phi_{\mathrm{f}}-\Phi_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}\right|
Solve and evaluate Inserting the appropriate values, we find:
\varepsilon_{\mathrm{in}}=(100)\left|\frac{\left(0.10 \mathrm{~T} \cdot \mathrm{m}^2\right)-0}{0.50 \mathrm{~s}-0}\right|=20 \mathrm{~V}
The magnitude of this emf is about what is required by laptop computers. However, the computer only works while the coil is turning. Laptops typically have a power requirement of about 50 watts (50 joules each second), which means considerable strength and endurance would be needed to keep the coil rotating.
Try it yourself: Determine the average emf produced in the coil if you turn the coil one quarter-turn in 1.0 s instead of in 0.50 s.
Answer: 10 V.