Motional Electromotive Force

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   The motional emf is given by $\varepsilon = (\vec{v}\times\vec{B})\cdot\vec{l}$. Since the rod is kept parallel to $\vec{B}$ and is also moved along the direction of $\vec{B}$, the velocity $\vec{v}$, field $\vec{B}$ and length $\vec{l}$ are all along the same direction. Induced emf, $\varepsilon = 0$, as $\vec{l}$, $\vec{B}$ and $\vec{v}$ are all in the same direction (so $\vec{v}\times\vec{B} = 0$).

2. 2.↑ back to the question

   Consider a small element $dx$ of the rod at a distance $x$ from the centre $O$. Its linear speed is $v = \omega x$, so the emf induced across this element is $d\varepsilon = (\vec{v}\times\vec{B})\,dx = vB\,dx = B\omega x\,dx$ The rod extends from $x = -l/2$ (end P) to $x = +l/2$ (end Q), with $O$ at $x=0$. The total emf induced across the rod, from P to Q, is $\varepsilon = \displaystyle\int_{-l/2}^{l/2} d\varepsilon = \int_{-l/2}^{l/2} B\omega x\,dx = \dfrac{B\omega}{2}\Big[x^2\Big]_{-l/2}^{l/2} = 0$ So the potential difference developed between the ends P and Q of the rod is zero. Note: this does NOT mean there is no induced emf anywhere in the rod — each half of the rod (from O to P, and from O to Q) individually has an emf of magnitude $B\omega l^2/8$ induced across it (since $O$ is at the highest potential, being the point of zero velocity... rather, being equidistant from both ends). It is only because the two halves are symmetric and contribute equal and opposite emfs from P to Q (going through the rod) that the net potential difference between the two ends P and Q comes out to be zero.

3. 3.↑ back to the question

   The self-induced emf in a coil is given by $\varepsilon = -L\dfrac{dI}{dt}$, where $L$ is the self-inductance of the coil (a constant for a given coil). So $|\varepsilon|$ is directly proportional to $\dfrac{dI}{dt}$, and the graph of induced emf versus rate of change of current is a straight line passing through the origin (with slope of magnitude $L$).

4. 4.↑ back to the question

   $\varepsilon = Blv\sin\theta$ Given: $B = 0.3\times 10^{-4}\text{ Wb m}^{-2}$, $v = 5\text{ m s}^{-1}$, $l = 10\text{ m}$, $\theta = 90°$ (rod falls at right angles to the field) $\varepsilon = Blv\sin 90° = 0.3\times 10^{-4}\times 10\times 5\times 1$ $\varepsilon = 15\times 10^{-4}\text{ V} = 1.5\times 10^{-3}\text{ V} = 1.5\text{ mV}$

5. 5.↑ back to the question

   Length of each blade (= radius), $l = r = 0.5\text{ m}$; frequency, $n = 2\text{ rps}$; $B = 8\times 10^{-5}\text{ T}$. As the fan rotates, the linear velocity at the outer end of the blade is $r\omega$, and at the inner end (hinge) is $0$. $\therefore$ Average linear velocity, $v = \dfrac{0 + r\omega}{2} = \dfrac{1}{2}r\omega$ The emf induced across each blade is $e = Bvl = B\left(\dfrac{1}{2}r\omega\right)l = B\left(\dfrac{r}{2}(2\pi n)\right)l = B r \pi n l$ $e = 8\times 10^{-5}\times 0.5\times \dfrac{22}{7}\times 2\times 0.5 = 12.57\times 10^{-5}\text{ Volt}$

6. 6.↑ back to the question

   Area of loop, $A = (10\times 10^{-2})^2 = 10^{-2}\text{ m}^2$; velocity, $v = 8\text{ cm/s} = 8\times 10^{-2}\text{ m/s}$ $\dfrac{dB}{dx} = 10^{-3}\text{ T cm}^{-1} = 10^{-3}\times 10^2\text{ T m}^{-1} = 10^{-1}\text{ T m}^{-1}$ Rate of change of magnetic flux due to motion of the loop in the non-uniform field is $\phi_1 = A\left(\dfrac{dB}{dx}\right)\left(\dfrac{dx}{dt}\right) = A\left(\dfrac{dB}{dx}\right)v$ $\phi_1 = 10^{-2}\times 10^{-1}\times 8\times 10^{-2} = 8\times 10^{-5}\text{ Wb s}^{-1}$ $\therefore$ Induced emf, $e = 8\times 10^{-5} = 8\times 10^{-5}\text{ V}$

7. 7.↑ back to the question

   (i) Induced emf, $\varepsilon = Blv = 0.1\times 0.1\times 20 = 0.2\text{ V}$ (ii) Since resistances of MN, NP and MQ are negligible compared to the resistance of the movable arm PQ, the overall resistance of the rectangular loop remains the same as that of PQ, i.e. $2\,\Omega$. Current, $I = \dfrac{\varepsilon}{R} = \dfrac{0.2}{2} = 0.1\text{ A}$

8. 8.↑ back to the question

   (a) Let the rod MN of length $l$ move with uniform velocity $\vec{v}$ on two parallel rails, in the presence of a uniform magnetic field $B$ directed into the page. In time $dt$, the rod moves a distance $dx$, so the change in area of the loop is $dA = l\,dx$. Change in flux, $d\phi = B\,dA = Bl\,dx$, so induced emf $\varepsilon = \left|\dfrac{d\phi}{dt}\right| = Bl\dfrac{dx}{dt} = Blv$ So induced current flows through the circuit: $i = \dfrac{\varepsilon}{R} = \dfrac{Blv}{R}$, where $R (= r)$ is the resistance of the rails (here, the moving rod MN). By Lenz's law (or Fleming's right-hand rule), this induced current opposes the increase in flux and flows in the sense that produces a retarding force on MN. (b) The retarding (opposing) force exerted on the rod by the induced current is $\vec{F}_m = i(\vec{l}\times\vec{B})$, i.e. $F_m = Bil = \dfrac{B^2l^2v}{R}$ For the rod to move with uniform velocity, an external force equal and opposite to this must be applied: $F_{ext} = F_m = \dfrac{B^2l^2v}{R}$ Power required to move the arm (mechanical power delivered by the external agent) is $P = F_{ext}\cdot v = \dfrac{B^2l^2v^2}{R}$

9. 9.↑ back to the question

   (i) We know $I = \dfrac{dq}{dt} \Rightarrow dq = I\,dt$ $\therefore q = \displaystyle\int I\,dt = \text{area under the } I\text{-}t\text{ graph} = \dfrac{1}{2}\times 0.4\times 1.0 = 0.2\text{ C}$ (ii) As we know, $|\varepsilon| = \dfrac{d\phi}{dt} \Rightarrow dq\times R = d\phi$ (since $I = \varepsilon/R = (d\phi/dt)/R$) $\Rightarrow \phi = qR = 0.2\times 10 = 2\text{ Wb}$ (iii) Now, $\phi = B\cdot A = B(10\times 10^{-4})$ $\Rightarrow B = \dfrac{2}{10\times 10^{-4}} = 2000\text{ T}$

10. 10.↑ back to the question

    Induced emf, $\varepsilon = (\vec{v}\times\vec{B})\cdot l$. Consider a small element $dx$ of the rod at distance $x$ from the hinged (fixed) end. For the element, induced emf $d\varepsilon = Bv\,dx = B\omega x\,dx$ $\therefore \varepsilon = \displaystyle\int_0^l d\varepsilon = \int_0^l B\omega x\,dx = \dfrac{B\omega l^2}{2}$ $\Rightarrow \varepsilon \propto \omega$ and $\varepsilon \propto l^2$ (i) The graph of $\varepsilon$ versus $\omega$ (for fixed $l$) is a straight line through the origin. (ii) The graph of $\varepsilon$ versus $l$ (for fixed $\omega$) is a parabola (opening upward) through the origin, since $\varepsilon \propto l^2$.

11. 11.↑ back to the question

    Given $l = 20\text{ cm} = 0.2\text{ m}$, $B = 0.1\text{ T}$, $v = 10\text{ cm/s} = 0.1\text{ m/s}$ (i) While the loop is entering the field, magnetic flux through the loop is $\phi = B\cdot A = Blx$ (where $x$ is the length of the loop already inside the field), so $\phi$ rises linearly from 0 to a maximum $\phi_{max} = Bl\times l = 0.1\times 0.2\times 0.2 = 0.004\text{ Wb} = 4\times 10^{-3}\text{ Wb}$ The flux $\phi$ then stays constant at $\phi_{max}$ while the loop is fully inside the field (since the field region, at 80 cm, is wider than the 20 cm loop), and finally decreases linearly back to zero as the loop exits the field on the far side. The $\phi$–$t$ graph is thus trapezoidal: rising linearly, flat at the top, then falling linearly, symmetric about the midpoint of the motion. (ii) Induced emf, $\varepsilon = -\dfrac{d\phi}{dt} = -Blv$ (while flux is changing) $|\varepsilon|_{max} = 0.1\times 0.2\times 0.1 = 0.002\text{ V} = 2\times 10^{-3}\text{ V}$ So the $\varepsilon$–$t$ graph consists of a constant negative block of value $-2\times 10^{-3}\text{ V}$ while the loop is entering the field (flux increasing), zero while the loop is fully inside (flux constant), and a constant positive block of $+2\times 10^{-3}\text{ V}$ while the loop is exiting the field (flux decreasing). (iii) Induced current, $I = \dfrac{|\varepsilon|}{R} = \dfrac{2\times 10^{-3}}{0.1} = 2\times 10^{-2}\text{ A}$ The $I$–$t$ graph has exactly the same shape as the $\varepsilon$–$t$ graph (just scaled by $1/R$): $-2\times 10^{-2}$ A while entering, zero while fully inside, and $+2\times 10^{-2}$ A while exiting the field.