Question

a 15.0 - cm - diameter wire coil is perpendicular to a magnetic field 0.60 t pointing up. in 0.25 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. ε =

Explanation:

Step1: Calculate the area of the coil

The diameter \(d = 15.8\space cm=0.158\space m\), so the radius \(r=\frac{d}{2}=0.079\space m\). The area of a circle \(A=\pi r^{2}\), so \(A = \pi\times(0.079)^{2}\space m^{2}\approx0.0196\space m^{2}\)

Step2: Use the formula for induced emf

The formula for average induced emf is \(\mathcal{E}=-\frac{N\Delta\Phi}{\Delta t}\). Assuming \(N = 1\) (not given, but if not specified, single - turn coil is a common assumption in basic problems). The magnetic flux \(\Phi=BA\). The change in magnetic field \(\Delta B=B_{f}-B_{i}\), where \(B_{i} = 0.60\space T\) (up, take as positive) and \(B_{f}=- 0.15\space T\) (down). So \(\Delta B=-0.15 - 0.60=-0.75\space T\). \(\Delta t = 0.25\space s\)

\(\mathcal{E}=\frac{\vert\Delta\Phi\vert}{\Delta t}=\frac{\vert\Delta B\vert A}{\Delta t}\)

Substitute \(\vert\Delta B\vert = 0.75\space T\), \(A\approx0.0196\space m^{2}\) and \(\Delta t = 0.25\space s\)

\(\mathcal{E}=\frac{0.75\times0.0196}{0.25}\space V\)

\(\mathcal{E}=0.0588\space V\approx0.059\space V\)

Answer:

\(\mathcal{E} = 0.059\space V\)