6.7 Energy Consideration: A Quantitative Study
In Section 6.5, we discussed qualitatively that Lenz’s law is consistent with the law of conservation of energy. Now we shall explore this aspect further with a concrete example.
Let r be the resistance of movable arm PQ of the rectangular conductor shown in Fig. 6.10. We assume that the remaining arms QR, RS and SP have negligible resistances compared to r. Thus, the overall resistance of the rectangular loop is r and this does not change as PQ is moved. The current I in the loop is,
\[ \begin{aligned} I &=\frac{\varepsilon}{r} \\ &=\frac{B l v}{r} \end{aligned} \]
On account of the presence of the magnetic field, there will be a force on the arm PQ. This force \( I(\mathbf{1} \times \mathbf{B}) \), is directed outwards in the direction opposite to the velocity of the rod. The magnitude of this force is,
\[ F=I l B=\frac{B^{2} l^{2} v}{r} \]
where we have used Eq. (6.7). Note that this force arises due to drift velocity of charges (responsible for current) along the rod and the consequent Lorentz force acting on them.
Alternatively, the arm PQ is being pushed with a constant speed v, the power required to do this is,
\[ \begin{aligned} P &=F v \\ &=\frac{B^{2} l^{2} v^{2}}{r} \end{aligned} \]
The agent that does this work is mechanical. Where does this mechanical energy go? The answer is: it is dissipated as Joule heat, and is given by the below which is identical to Eq. (6.8).
\[ P_{J}=I^{2} r=\left(\frac{B l v}{r}\right)^{2} r=\frac{B^{2} l^{2} v^{2}}{r} \]
Thus, mechanical energy which was needed to move the arm PQ is converted into electrical energy (the induced emf) and then to thermal energy. There is an interesting relationship between the charge flow through the circuit and the change in the magnetic flux. From Faraday’s law, we have learnt that the magnitude of the induced emf is,
\[ |\varepsilon|=\frac{\Delta \Phi_{\mathrm{B}}}{\Delta t} \] However, \[ |\varepsilon|=I r=\frac{\Delta Q}{\Delta t} r \] Thus, \[ \Delta Q=\frac{\Delta \Phi_{\mathrm{B}}}{r} \]