Question

A 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.9-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.15 s.What is the average induced emf in the loop?

Answer 1

Answer:

0.3405V

Explanation:

#Given a magnetic field of $1.9T$, diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:

$\phi=B.(\pi r^2)cos 0\textdegree\\=1.9\pi\times 0.0925^2 \times cos 0\textdegree\\=5.107\times10^-^2 \ Tm^2$

we can now calculate the induced emf, $\frac{\phi}{\bigtriangleup t}$:

$\frac{\phi}{\bigtriangleup t}=\frac{5.107\times 10^-^2}{0.15}\\=3.405\times 10^-^1V$

Hence, the induced emf of the loop is 0.3405V