Question

A conducting rod of length $l$ is moving perpendicular to the magnetic field. The rod moves from $0to2b$ while the field exists only from $0tob$. Find the graph for emf and power dissipated with respect to $x$.

Answer 1

Answer:

Calculating EMF induced:

• From the given figure it is clear that the field exists from $0tob$ only, therefore, the given conductor will experience a field only from $0tob$.
• As long as field exists a constant emf$\left(\epsilon \right)$ will be induced in the conductor.
• Induced emf in the conductor,$\epsilon =Blv$
• Due to this emf current developed in the conductor as,$i=\frac{\epsilon }{R}$

$\Rightarrow i=\frac{Blv}{R}$

Calculating Power dissipation:

Now, Power dissipation exists as long as emf exists in the conductor, $P=i^{2}R$

$$\begin{gathered} \Rightarrow P=\frac{{\displaystyle \raisebox{1ex}{${\left(Blv\right)}^2$}\!\left/ \!\raisebox{-1ex}{$R^2$}\right.}}{R} \\ \therefore P=\frac{{\displaystyle {\left(Blv\right)}^2}}{R^3} \end{gathered}$$

Hence, the graph of $\mathit{\epsilon }\mathbf{}\mathit{v}\mathit{s}\mathbf{}\mathit{x}$ and $\mathit{p}\mathbf{}\mathit{v}\mathit{s}\mathbf{}\mathit{x}$ is as follows,