Inductance

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

   (i) $M = \dfrac{\phi}{I_1} = \dfrac{15 - 10}{6 - 4} = \dfrac{5}{2} = 2.5\text{ Henry}$ (ii) $e = M\dfrac{dI_1}{dt}$ $\dfrac{dI_1}{dt} = \dfrac{e}{M} = \dfrac{100}{2.5} = 40\text{ A/s}$

2. 2.↑ back to the question

   Since $r_1 > r_2$ (coil Y lies well within coil X), when a current $I$ flows through the outer coil X, the magnetic field at the centre — and, since $r_2$ is small compared to $r_1$, essentially uniform over the entire area of the inner coil Y — is $B = \dfrac{\mu_0 N_1 I}{2r_1}$ The magnetic flux linked with coil Y (of area $\pi r_2^2$ and $N_2$ turns) is $N_2\phi_2 = B\times \pi r_2^2\times N_2 = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}I$ (i) $\therefore$ Mutual inductance, $M = \dfrac{N_2\phi_2}{I} = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}$ (ii) Magnetic flux linked with Y is $N_2\phi_2 = MI = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}I$

3. 3.↑ back to the question

   (a) The phenomenon of inducing an emf in a circuit by changing the magnetic flux in a neighbouring circuit is called mutual induction. The S.I. unit of mutual inductance is the henry (H). (b) Consider two straight, infinitely long, parallel conducting wires carrying currents $I_1$ and $I_2$, separated by a distance $x$. The force per unit length between them is $\dfrac{F}{a} = \dfrac{\mu_0 I_1 I_2}{2\pi x}$ The near side of the loop (at distance $x$, carrying current $I_2$ parallel to $I_1$) experiences a force of attraction: $F_1 = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2 a}{x}$ The far side of the loop (at distance $x+a$, carrying current $I_2$ antiparallel to $I_1$) experiences a force of repulsion: $F_2 = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2 a}{x+a}$ The forces on the two sides of the loop perpendicular to the wire cancel out by symmetry. Net force on the loop, $F_{net} = F_1 - F_2 = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2 a^2}{x(x+a)}$, directed towards the wire (i.e. the loop is attracted towards the wire, since the near-side attraction exceeds the far-side repulsion).

4. 4.↑ back to the question

   Mutual inductance: the phenomenon of inducing an emf in a circuit by changing the current or magnetic flux in a neighbouring circuit is called mutual induction, and the constant of proportionality $M$ between the flux linked with one coil and the current in the other is called their mutual inductance. Consider two long coaxial solenoids, each of length $l$: let $S_1$ (inner, radius $r_1$) have $n_1$ turns per unit length, and $S_2$ (outer, radius $r_2$) have $n_2$ turns per unit length, with total turns $N_1$ and $N_2$ respectively. When a current $I$ is passed through the outer solenoid $S_2$, the magnetic field produced inside it is $B_2 = \mu_0 n_2 I$ The flux linked with the inner solenoid $S_1$ is $N_1\phi_1 = B_2(\pi r_1^2)n_1 l = \mu_0 n_1 n_2 \pi r_1^2 lI$ (using the inner solenoid's cross-sectional area, since that is the area common to both) $\therefore M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l$ Similarly, when a current $I$ is passed through the inner solenoid $S_1$, the flux linked with the outer solenoid $S_2$ gives $M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l$ So $M_{12} = M_{21} = M$. Hence, the coefficient of mutual induction between the two coaxial solenoids is $M = \mu_0 n_1 n_2 \pi r_1^2 l$, or equivalently $M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$

5. 5.↑ back to the question

   (i) The phenomenon of inducing a current (emf) in a circuit by changing the current or flux in a neighbouring circuit is called mutual induction. The S.I. unit of mutual inductance is the henry (H). (ii) Here, $M = 1.5$ H, $\Delta I_1 = 20$ A, $\Delta t = 0.5$ s, $\Delta\phi_2 = ?$ We know, emf induced in the second coil, $\varepsilon = -\dfrac{(\Delta\phi)_2}{\Delta t} = -M\dfrac{\Delta I_1}{\Delta t}$ $\therefore (\Delta\phi)_2 = M\Delta I_1 = 1.5\times 20 = 30\text{ Wb}$

6. 6.↑ back to the question

   (i) Given $r_1 = 1\text{ cm} = 1\times 10^{-2}\text{ m}$ (inner loop) and $r_2 = 20\text{ cm} = 20\times 10^{-2}\text{ m}$ (outer loop). The mutual inductance of two concentric coplanar coils (single-turn, $N_1 = N_2 = 1$) is given by $M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{2r_2}$ $M = \dfrac{4\pi\times 10^{-7}\times 1\times 1\times \pi(1\times 10^{-2})^2}{2(20\times 10^{-2})} = \pi^2\times 10^{-10}\text{ H}$ (ii) Given $\dfrac{dI}{dt} = 5\text{ A/ms} = 5\times 10^3\text{ A/s}$ $|\varepsilon| = M\dfrac{dI}{dt} = \pi^2\times 10^{-10}\times 5\times 10^3 = 4.93\times 10^{-6}\text{ V} = 4.93\,\mu\text{V}$

7. 7.↑ back to the question

   Mutual inductance of a pair of coils is defined as the emf induced in one of the coils when the rate of change of current in the other coil is unity. When current $I_2$ flows through the outer coil-2 (radius $r_2$), the magnetic field produced at its centre is $B_2 = \dfrac{\mu_0 I_2}{2r_2}$, directed normal to the plane of the coils. Since $r_1 < r_2$, this field may be taken as uniform over the area of the inner coil-1, so the flux linked with coil-1 is $\phi_{12} = B_2 A_1\cos 0° = \dfrac{\mu_0 I_2}{2r_2}\times \pi r_1^2\times 1$ $\therefore \dfrac{\phi_{12}}{I_2} = \dfrac{\mu_0\pi r_1^2}{2r_2}$, so the mutual inductance is $M_{12} = \dfrac{\mu_0\pi r_1^2}{2r_2}$

8. 8.↑ back to the question

   (a) Mutual inductance is the phenomenon of an opposing emf being produced in one coil as a result of a change in current (or flux) in a neighbouring coil. S.I. unit is henry (H). (b) [Same derivation as for two coaxial solenoids of equal length above:] Considering an inner solenoid $S_1$ (radius $r_1$, $n_1$ turns per unit length) and outer solenoid $S_2$ (radius $r_2$, $n_2$ turns per unit length), each of length $l$: $M = \mu_0 n_1 n_2 \pi r_1^2 l = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$ (c) The mutually induced emf in coil $C_1$ (as capacitor $C_2$'s current changes) is $\varepsilon = -\dfrac{d\phi}{dt} = -\dfrac{d}{dt}(Mi) = -M\dfrac{di}{dt}$, where $\dfrac{di}{dt}$ is the rate of change of current in the neighbouring coil $C_2$.

9. 9.↑ back to the question

   The magnetic flux $\phi_B$ linked with a secondary coil is directly proportional to the current $I$ flowing through the primary coil, i.e. $\phi_B \propto I$ or $\phi_B = MI$, where $M$ is a constant of proportionality called the mutual inductance of one circuit with respect to the other. Equivalently, the mutual inductance of a pair of coils is the emf induced in one coil when the rate of change of current in the neighbouring coil is unity. The value of $M$ depends upon: (i) the characteristics (size, shape, number of turns) of both the coils/circuits, and (ii) their relative orientation and position with respect to each other.

10. 10.↑ back to the question

    Consider two long coaxial solenoids, each of length $l$. Let $n_1$ be the number of turns per unit length of the inner solenoid $S_1$ (radius $r_1$), and $n_2$ the number of turns per unit length of the outer solenoid $S_2$ (radius $r_2$). Let $N_1$ and $N_2$ be the total number of turns of $S_1$ and $S_2$ respectively. When a current $I$ is passed through $S_2$, the field inside it is $B_2 = \mu_0 n_2 I$; the flux linked with $S_1$ is $N_1\phi_1 = B_2(\pi r_1^2)n_1 l$ (since flux threads only the common cross-sectional area, that of the narrower, inner solenoid $S_1$) $N_1\phi_1 = \mu_0 n_1 n_2 \pi r_1^2 lI \;\Rightarrow\; M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l$ Similarly, passing a current through $S_1$ and finding the flux linked with $S_2$ gives $M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l$ Since $M_{12} = M_{21} = M$, the mutual inductance of the two coaxial solenoids (of different radii and different turns) is $M = \mu_0 n_1 n_2 \pi r_1^2 l$, or equivalently $M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$ (note: it is the radius $r_1$ of the narrower/inner solenoid that appears here, regardless of which solenoid carries the current, by the reciprocity $M_{12}=M_{21}$)

11. 11.↑ back to the question

    Self-inductance, $L = \dfrac{\mu_0\mu_r N^2 A}{l}$. When $N' = 2N$ (length and area unchanged), $L' = \dfrac{\mu_0\mu_r A(2N)^2}{l} = \dfrac{4\mu_0\mu_r N^2 A}{l} = 4L$ Thus, the self-inductance of the coil becomes 4 times.

12. 12.↑ back to the question

    For an ideal inductor, $|\varepsilon| = L\dfrac{dI}{dt}$, so the graph showing the variation of the induced emf $|\varepsilon|$ with the rate of change of current $dI/dt$ is a straight line passing through the origin, with slope equal to the self-inductance $L$ of the inductor.

13. 13.↑ back to the question

    Self-inductance: when the current in a coil is changed, a back emf is induced in the same coil. This phenomenon is called self-inductance. If $L$ is the self-inductance of the coil, then $N\phi \propto I$, i.e. $N\phi = LI$ $\therefore L = \dfrac{N\phi}{I}$ The S.I. unit of self-inductance is the henry (H).

14. 14.↑ back to the question

    (a) Self-inductance: when the current in a coil is changed, a back emf is induced in the same coil; this is called self-inductance. If $L$ is the self-inductance of the coil, $N\phi = LI \Rightarrow L = \dfrac{N\phi}{I}$. The S.I. unit of self-inductance is the henry (H). (b) Let $S_1$ (inner solenoid, radius $r_1$, $n_1$ turns per unit length) and $S_2$ (outer solenoid, radius $r_2$, $n_2$ turns per unit length) be coaxial, each of length $L$. When current $I$ flows through the outer solenoid $S_2$, the field it produces is $B_2 = \mu_0 n_2 I$ The flux linked with the inner solenoid $S_1$ is $N_1\phi_1 = B_2(\pi r_1^2)n_1 L = \mu_0 n_1 n_2 \pi r_1^2 LI$ (using $S_1$'s cross-sectional area, being the smaller of the two) $\therefore M_{12} = \mu_0 n_1 n_2 \pi r_1^2 L$ By reciprocity, $M_{12} = M_{21} = M = \mu_0 n_1 n_2 \pi r_1^2 L$

15. 15.↑ back to the question

    Self-inductance: the phenomenon by which a coil opposes any change in current flowing through it, by inducing a back emf in itself, is called self-inductance. Consider a circuit with an inductor $L$ and resistor $R$ connected to a source of emf. As the current grows, part of the work done by the source is stored in the inductor as magnetic-field energy, and the rest appears as heat in the resistor. If $I$ is the current at time $t$, the self-induced emf is $\varepsilon = L\dfrac{dI}{dt}$ In a small time $dt$, work done by the source against this back emf is $dW = \varepsilon I\,dt = L\dfrac{dI}{dt}I\,dt = LI\,dI$ Total work done in building up the current from $0$ to $I_0$: $W = \displaystyle\int_0^{I_0} LI\,dI = \dfrac{1}{2}LI_0^2$ This entire work done is stored as magnetic potential energy in the inductor: $U = \dfrac{1}{2}LI^2$

16. 16.↑ back to the question

    (i) Self-inductance: when the current in a coil is changed, a back emf is induced in the same coil; this phenomenon is called self-inductance. $L = N\phi/I$. S.I. unit: henry (H). (ii) Here, $N_1 = 15$ (turns per cm, treating $l = 1\text{ cm} = 10^{-2}\text{ m}$), $N_2 = 1$ (the single loop), $A = 2.0\text{ cm}^2 = 2.0\times 10^{-4}\text{ m}^2$ Mutual inductance of the solenoid–loop system, $M = \dfrac{\mu_0 N_1 N_2 A}{l} = \dfrac{4\pi\times 10^{-7}\times 15\times 1\times 2.0\times 10^{-4}}{10^{-2}} = 120\pi\times 10^{-9}\text{ H}$ Induced emf in the loop, $\varepsilon_2 = M\dfrac{\Delta i_1}{\Delta t}$ (numerically) $= 120\pi\times 10^{-9}\times \dfrac{(4-2)}{0.1}$ $\varepsilon_2 = 120\times 3.14\times 10^{-9}\times \dfrac{2}{0.1} = 7.5\times 10^{-6}\text{ V} = 7.5\,\mu\text{V}$

17. 17.↑ back to the question

    (i) Induced voltage, $V = L\dfrac{di}{dt}$. Since $\dfrac{di}{dt}$ is the same for both coils, $\dfrac{V_1}{V_2} = \dfrac{L_1}{L_2} = \dfrac{16}{12} = \dfrac{4}{3}$ (ii) Power, $P = iV$. Since $P$ is the same (equal) for both coils, $\dfrac{i_1}{i_2} = \dfrac{V_2}{V_1} = \dfrac{3}{4}$ (iii) Energy stored, $E = \dfrac{1}{2}Li^2$ $\dfrac{E_1}{E_2} = \dfrac{L_1 i_1^2}{L_2 i_2^2} = \dfrac{16}{12}\times \dfrac{9}{16} = \dfrac{3}{4}$

18. 18.↑ back to the question

    (i) Length of conducting rod $XY = \lambda$, moving with velocity $v$, in magnetic field $B$. Since the rod is perpendicular to the field ($\theta = 90°$), the induced emf is $|\varepsilon| = B\lambda v\sin 90° = B\lambda v$ By the right-hand rule (or Fleming's right-hand rule) applied to the direction of $v$, $B$ and the rod, one end of the rod becomes the positive terminal and the other the negative terminal of this induced emf source. (ii) So, induced current flows through the circuit, $i = \dfrac{\varepsilon}{r} = \dfrac{B\lambda v}{r}$, where $r$ is the resistance connected between M and N. The direction of this current (in the external resistor $r$, from N to M or M to N) is determined by Lenz's law / the right-hand rule, such that it opposes the change in flux caused by the motion of the rod.

19. 19.↑ back to the question

    (i) Suppose current $I$ is flowing through an inductor of self-inductance $L$. Then the magnetic flux linked with it is given by $\phi = LI$ Since $L$ is constant, the graph of magnetic flux $\phi$ versus current $I$ is a straight line through the origin (slope $= L$). (ii) Induced emf is given by $\varepsilon = -\dfrac{d\phi}{dt} = -\dfrac{d}{dt}(LI) = -L\dfrac{dI}{dt}$, so $|\varepsilon| = L\dfrac{dI}{dt}$ The graph of induced emf $|\varepsilon|$ versus $dI/dt$ is also a straight line through the origin (slope $= L$). (iii) Magnetic potential energy stored is $U = \dfrac{1}{2}LI^2$, so $U \propto I^2$ The graph of magnetic potential energy $U$ versus current $I$ is a parabola (opening upward) through the origin.

20. 20.Answer: (A)↑ back to the question

    Answer: (a) 50 V Change in current, $dI = 7 - 3 = 4$ A; time, $dt = 0.04$ s; mutual inductance, $M = 0.5$ H Induced emf, $e = M\dfrac{dI}{dt} = 0.5\times \dfrac{4}{0.04} = 50\text{ V}$

21. 21.Answer: (D)↑ back to the question

    Answer: (d) 75 mH Self-inductance, $L = 108$ mH for $N = 600$ turns. Let $L'$ be the self-inductance for $N' = 500$ turns. Since $L = \dfrac{\mu_0 N^2 A}{l}$, we have $L \propto N^2$, so $\dfrac{L'}{L} = \left(\dfrac{N'}{N}\right)^2 = \left(\dfrac{500}{600}\right)^2$ $L' = 108\times \dfrac{5^2}{6^2} = 108\times \dfrac{25}{36} = 75\text{ mH}$

22. 22.Answer: (C)↑ back to the question

    Answer: (c) The rate of heating When the iron rod (of high relative permeability $\mu_r$) is inserted into the solenoid, the magnetic field at the centre, the magnetic flux linked with the solenoid, and the self-inductance $L = \mu_0\mu_r n^2 Al$ of the solenoid all increase. However, since the current through the solenoid is held CONSTANT, the rate of Joule heating, $I^2R$, depends only on the current and the (unchanged) resistance of the winding — it does not depend on $L$ or $B$, so it does not change.