Question

a coil has 250 turns enclosing an area of 11.2 cm². in a physics laboratory experiment, the coil is rotated during the time interval 0.042 s from a position in which the plane of each turn is perpendicular to earths magnetic field to one in which the plane of each turn is parallel to the field. the magnitude of earths magnetic field at the lab location is 5.40×10⁻⁵ t. what is the magnitude of the average emf induced in the entire coil? express your answer in volts. view available hint(s) |ε| = v

Explanation:

Step1: Calculate the change in magnetic flux

The formula for magnetic flux is $\varPhi = BA\cos\theta$. Initially, $\theta_{1} = 0^{\circ}$ (plane perpendicular to field, so $\cos\theta_{1}=1$), and finally $\theta_{2} = 90^{\circ}$ (plane parallel to field, so $\cos\theta_{2}=0$). The change in flux $\Delta\varPhi=\varPhi_{2}-\varPhi_{1}=B A(\cos\theta_{2}-\cos\theta_{1})$. Given $B = 5.40\times 10^{-5}\text{ T}$, $A=11.2\text{ cm}^2=11.2\times 10^{- 4}\text{ m}^2$. Then $\Delta\varPhi=(5.40\times 10^{-5}\text{ T})\times(11.2\times 10^{-4}\text{ m}^2)\times(0 - 1)=-6.048\times 10^{-8}\text{ Wb}$

Step2: Use Faraday's law of electromagnetic induction

Faraday's law is $\mathcal{E}=-N\frac{\Delta\varPhi}{\Delta t}$. Given $N = 250$ turns and $\Delta t=0.042\text{ s}$. Substitute the values: $\mathcal{E}=-250\times\frac{- 6.048\times 10^{-8}\text{ Wb}}{0.042\text{ s}}$

First, calculate $\frac{-6.048\times 10^{-8}}{0.042}=-1.44\times 10^{-6}$. Then $-250\times(-1.44\times 10^{-6}) = 3.6\times 10^{-4}\text{ V}$

Answer:

$3.6\times 10^{-4}\text{ V}$