Energy Consideration: A Quantitative Study

The question this section answers

• A motional emf drives a current, and the current dissipates heat. Where does that energy come from? The answer is that it comes entirely from the external agent that keeps the conductor moving — which is Lenz's law expressed as a book-keeping identity.

Set-up and diagram in words

• A rectangular conducting frame lies in the plane of the page in a uniform magnetic field $\vec{B}$ directed INTO the page. One arm, of length $l$ and resistance $R$ (or $r$), can slide on two frictionless parallel rails of negligible resistance; the far ends of the rails are joined, closing the circuit.
• The movable arm is pulled with a constant velocity $\vec{v}$ perpendicular to itself. Mark on the figure: $\vec{B}$ into the page, $\vec{v}$ along the rails, the length $l$, the induced current $i$ round the loop, and the retarding force $\vec{F}_m$ on the arm, opposite to $\vec{v}$.

The derivation, step by step

• In time $dt$ the arm moves a distance $dx$, so the change of area is $dA = l\,dx$ and the change of flux is $d\phi = B\,dA = Bl\,dx$.
• Induced emf:

$$\varepsilon = \left\lvert\dfrac{d\phi}{dt}\right\rvert = Bl\dfrac{dx}{dt} = Blv$$

• Induced current, with $R$ the total resistance of the circuit (usually the movable arm alone, the rails being of negligible resistance):

$$i = \dfrac{\varepsilon}{R} = \dfrac{Blv}{R}$$

• By Lenz's law (or Fleming's right-hand rule) this current flows in the sense that opposes the change of flux, and hence produces a RETARDING force on the arm.
• Force on the current-carrying arm in the field:

$$\vec{F}_m = i(\vec{l}\times\vec{B}),\qquad F_m = Bil = \dfrac{B^2l^2v}{R}$$

directed opposite to $\vec{v}$.

• For the arm to keep moving with UNIFORM velocity, the external agent must apply an equal and opposite force:

$$F_{\text{ext}} = F_m = \dfrac{B^2l^2v}{R}$$

• Power supplied by the external agent:

$$P_{\text{mech}} = F_{\text{ext}}\,v = \dfrac{B^2l^2v^2}{R}$$

Energy balance — the point of the section

• Power dissipated as Joule heat in the circuit:

$$P_{\text{Joule}} = i^2R = \left(\dfrac{Blv}{R}\right)^{2}R = \dfrac{B^2l^2v^2}{R}$$

$$\therefore\quad P_{\text{mech}} = P_{\text{Joule}}$$

• Every joule of mechanical work done against the magnetic force reappears as electrical energy and is finally dissipated as heat. Nothing is created and nothing is lost. This is why Lenz's law must take the form it does.
• Charge that flows while the arm sweeps a distance $\Delta x$:

$$q = \int i\,dt = \dfrac{\Delta\phi}{R} = \dfrac{Bl\,\Delta x}{R}$$

• Important qualification CBSE tests: once the frame has moved FULLY inside the field region, or fully outside it, the flux stops changing, so $i=0$, $F_m=0$ and no external force or power is needed at all. The retarding force exists only while the frame is partly in and partly out.

Numericals set on this section

• Frame MNOP of resistance $R$, arm ON of length $l$ inside the field, pulled out with velocity $v$: force on ON is $F_{ON} = \dfrac{B^2l^2v}{R}$, opposing $\vec{v}$; power required $P = \dfrac{B^2l^2v^2}{R}$; and $F_{ON}=0$ once the frame is fully in or fully out.
• Rod PQ of length $20\ \text{cm}$ and resistance $0.1\ \Omega$ on short-circuited smooth rails in $B = 0.4\ \text{T}$, moved at $v = 10\ \text{cm s}^{-1}$:

$$F = \dfrac{B^2l^2v}{R} = \dfrac{(0.4)^2\times(0.2)^2\times(0.1)}{0.1} = 6.4\times10^{-3}\ \text{N}$$

$$P = Fv = 6.4\times10^{-3}\times 0.1 = 6.4\times10^{-4}\ \text{W}$$

The rotating rod, done quantitatively

• A rod of length $l$ and resistance $R$ is rotated at frequency $\nu$ with one end hinged at the centre of a circular metallic ring of radius $l$, the other end touching the ring, in a uniform field $B$ parallel to the axis.
• In time $t$ the rod turns through $\theta$, sweeping an area $A = \pi l^2\times\dfrac{\theta}{2\pi} = \dfrac{1}{2}l^2\theta$, so

$$\phi = B\left(\dfrac{1}{2}l^2\theta\right)\cos 0^{\circ} = \dfrac{1}{2}Bl^2\theta$$

$$\varepsilon = \dfrac{d\phi}{dt} = \dfrac{1}{2}Bl^2\dfrac{d\theta}{dt} = \dfrac{1}{2}Bl^2\omega$$

$$i = \dfrac{\varepsilon}{R} = \dfrac{Bl^2\omega}{2R}$$

• Force on the rod carrying this current in the field, using $\omega = 2\pi\nu$:

$$F = ilB = \dfrac{Bl^2\omega}{2R}\cdot lB = \dfrac{\pi\nu B^2l^3}{R}$$

• By Lenz's law this force opposes the rotation, so the external agent must apply an equal and opposite force to maintain constant $\omega$.
• Power required, with $v = l\omega = 2\pi l\nu$ the speed of the free end:

$$P = Fv = \dfrac{\pi\nu B^2l^3 v}{R}$$

Standard results

• $\varepsilon = Blv$, $i = \dfrac{Blv}{R}$, $F = \dfrac{B^2l^2v}{R}$, $P = \dfrac{B^2l^2v^2}{R} = i^2R$.
• $q = \dfrac{Bl\,\Delta x}{R}$, independent of the speed.
• Rotating rod: $\varepsilon = \dfrac{1}{2} Bl^2\omega$, $i = \dfrac{Bl^2\omega}{2R}$, $F = \dfrac{\pi\nu B^2l^3}{R}$, $P = Fv$ with $v = 2\pi l\nu$.
• Mechanical power in $=$ electrical power out $=$ Joule heat: state this line explicitly, it carries a mark.