Question

a fixed 18.6 - cm - diameter wire coil is perpendicular to a magnetic field 0.69 t pointing up. in 0.26 s, the field is changed to 0.15 t pointing down. part a what is the average induced emf in the coil? express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Calculate the area of the coil

The diameter \(d = 18.6\space cm=0.186\space m\), so the radius \(r=\frac{d}{2}=\frac{0.186}{2}\space m = 0.093\space m\). The area of a circle \(A=\pi r^{2}\), so \(A=\pi\times(0.093)^{2}\space m^{2}\approx0.0271\space m^{2}\).

Step2: Determine the change in magnetic flux

The initial magnetic field \(B_{1} = 0.69\space T\) (up, let's take up as positive), the final magnetic field \(B_{2}=- 0.15\space T\) (down). The change in magnetic flux \(\Delta\Phi=\Delta B\times A=(B_{2}-B_{1})A=(-0.15 - 0.69)\times0.0271\space Wb=-0.022764\space Wb\).

Step3: Calculate the average induced emf

According to Faraday's law of electromagnetic induction \(\mathcal{E}=-\frac{\Delta\Phi}{\Delta t}\). Here \(\Delta t = 0.26\space s\). So \(\mathcal{E}=-\frac{- 0.022764}{0.26}\space V\approx0.0876\space V\approx0.088\space V\) (rounded to two significant figures).

Answer:

\(0.088\space V\)