Question
A flexible wire loop in the shape of a circle has radius that grown linearly with time. There is a magnetic field perpendicular to the plane of the loop that has a magnitude inversely proportional to the distance from the center of the loop, $$B\left( r \right) \propto \frac{1}{r}.$$ How does the emf $$E$$ vary with time?
A.
$$E \propto {t^2}$$
B.
$$E \propto t$$
C.
$$E \propto \sqrt t $$
D.
$$E$$ is constant
Answer :
$$E$$ is constant
Solution :
Let radius of the loop is $$r$$ at any time $$t$$ and in further time $$dt,$$ radius increases by $$dr.$$
The change in flux : $$d\phi = \left( {2\pi rdr} \right)B$$
$$\eqalign{ & \Rightarrow e = \frac{{d\phi }}{{dt}} = 2\pi r\left( {\frac{{dr}}{{dt}}} \right)\frac{k}{r} \cr & \Rightarrow e = 2\pi ck\left( {{\text{constant}}} \right)\,\,\left[ {\because \frac{{dr}}{{dt}} = c,B = \frac{k}{r}} \right] \cr} $$
The change in flux : $$d\phi = \left( {2\pi rdr} \right)B$$
Let radius of the loop is $$r$$ at any time $$t$$ and in further time $$dt,$$ radius increases by $$dr.$$
The change in flux : $$d\phi = \left( {2\pi rdr} \right)B$$
$$\eqalign{ & \Rightarrow e = \frac{{d\phi }}{{dt}} = 2\pi r\left( {\frac{{dr}}{{dt}}} \right)\frac{k}{r} \cr & \Rightarrow e = 2\pi ck\left( {{\text{constant}}} \right)\,\,\left[ {\because \frac{{dr}}{{dt}} = c,B = \frac{k}{r}} \right] \cr} $$
The change in flux : $$d\phi = \left( {2\pi rdr} \right)B$$
