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NCERT Exemplar · Class 12 Physics Moving Charges and Magnetism

29 questions

Long Answer Questions 23–29 (part 5 of 5)

  1. Exercise 23

    A 100\displaystyle 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (Fig. 4.4\displaystyle 4.4). A mass 500g is added to the other arm to balance the weight of the coil. A current 4.9\displaystyle 4.9 A passes through the coil and a constant magnetic field of 0.2\displaystyle 0.2 T acting inward (in xz\displaystyle x z plane) is switched on such that only arm CD of length 1\displaystyle 1 cm lies in the field. How much additional mass 'm\displaystyle m' must be added to regain the balance? NCERT_Question_Class12_Physics_Exemplar_Ch4_Q23
    Only arm CD (length \(\displaystyle l = 1\ \mathrm{cm}\)) lies in the field, so only it feels a force.\[F = N I l B = 100 \times 4.9\ \mathrm{A} \times 10^{-2}\ \mathrm{m} \times 0.2\ \mathrm{T} = 0.98\ \mathrm{N} \]Current in CD runs from C to D, \(\displaystyle \mathbf{B}\) is into the page:\[\mathbf{F} = N I\,(\mathbf{l}\times\mathbf{B}) \ \Rightarrow\ \mathbf{F}\ \text{points downward} \]NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q23The coil is pulled down, so the extra mass goes on the other arm:\[m g = F \ \Rightarrow\ m = \frac{0.98\ \mathrm{N}}{9.8\ \mathrm{m\,s^{-2}}} = 0.1\ \mathrm{kg} = 100\ \mathrm{g} \]Answer: \(\displaystyle m = 100\ \mathrm{g}\), added on the arm opposite the coil.

  2. Exercise 24

    A rectangular conducting loop consists of two wires on two opposite sides of length l\displaystyle l joined together by rods of length d\displaystyle d. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance R\displaystyle R and the rods are of low resistance, which in turn are connected to a constant voltage source V0\displaystyle V_0. The loop is placed in uniform a magnetic field B\displaystyle \mathbf{B} at 45\displaystyle 45° to its plane. Find τ\displaystyle \boldsymbol{\tau}, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.
    NCERT’s answer
    \(\displaystyle F_1=i_1 l B=\frac{V_0}{R} l B \quad \tau_1=\frac{d}{2 \sqrt{2}} F_1=\frac{V_0 l d B}{2 \sqrt{2} R}\) \[F_2=i_2 l B=\frac{V_0}{2 R} l B \quad \tau_2=\frac{d}{2 \sqrt{2}} F_2=\frac{V_0 l d B}{4 \sqrt{2} R} \] Net torque \(\displaystyle \tau=\tau_1-\tau_2\) \[\tau=\frac{1}{4 \sqrt{2}} \frac{V_0 A B}{R} \] NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q24_ncert
    Same material and length, half the cross-section \(\displaystyle \Rightarrow\) the thin wire has resistance \(\displaystyle 2R\). The rods are equipotential, so the wires are in parallel across \(\displaystyle V_0\):\[I_1 = \frac{V_0}{R}, \qquad I_2 = \frac{V_0}{2R} \]Both currents run the same way, rod to rod. Take \(\displaystyle \mathbf{B}\) perpendicular to the wires. Its in-plane part \(\displaystyle B\cos45^\circ\) gives each wire a force normal to the loop; the normal part of \(\displaystyle \mathbf B\) and the rods give no torque about the axis.\[F_1 = I_1\, l\, B\cos45^\circ, \qquad F_2 = I_2\, l\, B\cos45^\circ \]NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q24The two forces act the same way, at \(\displaystyle d/2\) on opposite sides of the axis, so the torques subtract:\[\tau = (F_1 - F_2)\,\frac{d}{2} = \frac{(I_1 - I_2)\,l\,d\,B}{2\sqrt{2}} = \frac{V_0}{2R}\cdot\frac{l\,d\,B}{2\sqrt{2}} \]Answer: \(\displaystyle \tau = \dfrac{V_0\,B\,l\,d}{4\sqrt{2}\,R}\)

  3. Exercise 25

    An electron and a positron are released from (0\displaystyle 0, 0\displaystyle 0, 0\displaystyle 0) and (0,0,1.5R)\displaystyle (0,0,1.5 R) respectively, in a uniform magnetic field B=B0i^\displaystyle \mathbf{B}=\mathrm{B}_0 \hat{\mathbf{i}}, each with an equal momentum of magnitude p=eBR\displaystyle p=e B R. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?
    NCERT’s answer
    As \(\displaystyle \mathbf{B}\) is along the \(\displaystyle x\) axis, for a circular orbit the momenta of the two particles are in the \(\displaystyle y-z\) plane. Let \(\displaystyle \mathbf{p}_1\) and \(\displaystyle \mathbf{p}_2\) be the momentum of the electron and positron, respectively. Both of them define a circle of radius \(\displaystyle R\). They shall define circles of opposite sense. Let \(\displaystyle \mathbf{p}_1\) make an angle \(\displaystyle \theta\) with the \(\displaystyle y\) axis \(\displaystyle \mathbf{p}_2\) must make the same angle. The centres of the repective circles must be perpendicular to the momenta and at a distance \(\displaystyle R\). Let the center of the electron be at \(\displaystyle C e\) and of the positron at \(\displaystyle C p\). The coordinates of Ce is The coordinates of Ce is \[C e \equiv(0,-R \sin \theta, R \cos \theta) \] The coordinates of Cp is \[C p \equiv\left(0,-R \sin \theta, \frac{3}{2} \mathrm{R}-\mathrm{R} \cos \theta\right) \] The circles of the two shall not overlap if the distance between the two centers are greater than \(\displaystyle 2 R\). Let \(\displaystyle d\) be the distance between Cp and Ce. Then \(\displaystyle d^2=(2 \mathrm{R} \operatorname{Sin} \theta)^2+\left(\frac{3}{2} \mathrm{R}-2 \mathrm{R} \cos \theta\right)^2\) \[\begin{aligned} & =4 R^2 \operatorname{Sin}^2 \theta+\frac{9^2}{4} R-6 R^2 \cos \theta+4 R^2 \cos ^2 \theta \\ & =4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta \end{aligned} \] Since \(\displaystyle d\) has to be greater than \(\displaystyle 2 R\) \[\begin{aligned} & d^2>4 R^2 \\ & \Rightarrow 4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta>4 R^2 \\ & \Rightarrow \frac{9}{4}>6 \cos \theta \end{aligned} \] Or, \(\displaystyle \cos \theta<\frac{3}{8}\). NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q25_ncert
    Radius of each orbit, with \(\displaystyle B = B_0\):\[r = \frac{p}{eB} = \frac{eBR}{eB} = R \]A circle (not a helix) needs no momentum along \(\displaystyle \mathbf B\): \(\displaystyle p_x = 0\). Both particles then stay in the plane \(\displaystyle x = 0\) on circles of radius \(\displaystyle R\). Let the momentum directions in the \(\displaystyle yz\)-plane be angles \(\displaystyle \alpha\) (electron) and \(\displaystyle \beta\) (positron), measured from \(\displaystyle +y\) towards \(\displaystyle +z\). The centres are\[C_e = R(-\sin\alpha,\ \cos\alpha), \qquad C_p = R(\sin\beta,\ 1.5 - \cos\beta) \]NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q25Equal circles in one plane do not meet only if the centres are more than \(\displaystyle 2R\) apart:\[C_eC_p^{2} = R^2\left[4.25 + 2\cos(\alpha - \beta) - 3(\cos\alpha + \cos\beta)\right] > 4R^2 \]\[\Rightarrow\ 2\cos(\alpha - \beta) - 3(\cos\alpha + \cos\beta) + \tfrac{1}{4} > 0 \]For equal directions \(\displaystyle \alpha = \beta\) this reads \(\displaystyle \cos\alpha < \tfrac{3}{8}\); both along \(\displaystyle -y\) gives \(\displaystyle C_eC_p = 3.5R\).Answer: \(\displaystyle p_x = 0\) (momenta in the \(\displaystyle yz\)-plane) and \(\displaystyle 2\cos(\alpha-\beta) - 3(\cos\alpha+\cos\beta) + \tfrac14 > 0\).

  4. Exercise 26

    A uniform conducting wire of length 12a\displaystyle 12 a and resistance R\displaystyle R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a\displaystyle a; (ii) a square of sides a\displaystyle a and, (iii) a regular hexagon of sides a\displaystyle a. The coil is connected to a voltage source V0\displaystyle V_0. Find the magnetic moment of the coils in each case.
    NCERT’s answer
    Area: \(\displaystyle \mathrm{A}=\frac{\sqrt{3}}{4} a^2 \quad \mathrm{~A}=\mathrm{a}^2 \quad A=\frac{3 \sqrt{3}}{4} a^2\) Current \(\displaystyle I\) is same for all Magnetic moment \(\displaystyle m=n I A\) \[\therefore m=I a^2 \sqrt{3} \quad 3 a^2 I \quad 3 \sqrt{3} a^2 I \] (Note: \(\displaystyle m\) is in a geometric series) NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q26_ncert
    Same wire, same \(\displaystyle V_0\) across it, so the current is the same in all three coils:\[I = \frac{V_0}{R} \]Turns \(\displaystyle N\) and area \(\displaystyle A\) of one turn:\[\text{(i)}\ N = \frac{12a}{3a} = 4,\quad A = \frac{\sqrt{3}}{4}a^2 \]\[\text{(ii)}\ N = \frac{12a}{4a} = 3,\quad A = a^2 \]\[\text{(iii)}\ N = \frac{12a}{6a} = 2,\quad A = \frac{3\sqrt{3}}{2}a^2 \]\[m = N I A \]\[m_{\triangle} = 4\cdot\frac{V_0}{R}\cdot\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}\,V_0 a^2}{R} \]\[m_{\square} = 3\cdot\frac{V_0}{R}\cdot a^2 = \frac{3\,V_0 a^2}{R} \]\[m_{\text{hex}} = 2\cdot\frac{V_0}{R}\cdot\frac{3\sqrt{3}}{2}a^2 = \frac{3\sqrt{3}\,V_0 a^2}{R} \]Answer: \(\displaystyle m = \dfrac{\sqrt{3}\,V_0a^2}{R},\ \dfrac{3\,V_0a^2}{R},\ \dfrac{3\sqrt{3}\,V_0a^2}{R}\ (\mathrm{A\,m^2})\) for the triangle, square and hexagon, each along the coil's axis.

  5. Exercise 27

    Consider a circular current-carrying loop of radius R\displaystyle R in the x−y\displaystyle x-y plane with centre at origin. Consider the line intergral I(L)=∣∫−LLB.dl∣\displaystyle \mathfrak{I}(L)=\left|\int_{-L}^L \mathbf{B} . \mathbf{d l}\right| taken along z-axis.
    (a)
    Show that I\displaystyle \mathfrak{I} (L) monotonically increases with L\displaystyle L.
    (b)
    Use an appropriate Amperian loop to show that I(∞)=μ0I\displaystyle \mathfrak{I}(\infty)=\mu_0 I, where I\displaystyle I is the current in the wire.
    (c)
    Verify directly the above result.
    (d)
    Suppose we replace the circular coil by a square coil of sides R\displaystyle R carrying the same current I\displaystyle I. What can you say about I(L)\displaystyle \mathfrak{I}(L) and I(∞)\displaystyle \mathfrak{I}(\infty)?
    (a) On the axis the loop's field points along \(\displaystyle z\), with one sign everywhere:\[B_z = \frac{\mu_0 I R^2}{2\,(R^2 + z^2)^{3/2}} > 0 \]\[\mathfrak{I}(L) = \int_{-L}^{L} B_z\,dz \ \Rightarrow\ \frac{d\mathfrak{I}}{dL} = B_z(L) + B_z(-L) = \frac{\mu_0 I R^2}{(R^2 + L^2)^{3/2}} > 0 \](b) Close the \(\displaystyle z\)-axis segment with a semicircle of radius \(\displaystyle L\) in the \(\displaystyle xz\)-plane (\(\displaystyle x > 0\)); the wire pierces this half-disc once, at \(\displaystyle (R, 0, 0)\).NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q27\[\oint \mathbf{B}\cdot d\mathbf{l} = \int_{-L}^{L} B_z\,dz + \int_{\text{arc}} \mathbf{B}\cdot d\mathbf{l} = \mu_0 I \]For \(\displaystyle L \gg R\) the field is a dipole field, \(\displaystyle B \le \dfrac{\mu_0 (I\pi R^2)}{2\pi L^3}\), over arc length \(\displaystyle \pi L\):\[\left|\int_{\text{arc}} \mathbf{B}\cdot d\mathbf{l}\right| \le \pi L\cdot\frac{\mu_0 (I\pi R^2)}{2\pi L^3} = \frac{\pi\mu_0 I R^2}{2L^2} \to 0 \ \Rightarrow\ \mathfrak{I}(\infty) = \mu_0 I \](c) Direct integration:\[\mathfrak{I}(L) = \frac{\mu_0 I R^2}{2}\left[\frac{z}{R^2\sqrt{R^2 + z^2}}\right]_{-L}^{L} = \frac{\mu_0 I\,L}{\sqrt{R^2 + L^2}} \]\[\mathfrak{I}(\infty) = \lim_{L\to\infty}\frac{\mu_0 I}{\sqrt{1 + R^2/L^2}} = \mu_0 I \](d) The loop of (b) never used the coil's shape, so the limit stays. The square lies inside the circle, nearer the centre, so its centre field is larger:\[B_{sq}(0) = \frac{2\sqrt{2}\,\mu_0 I}{\pi R} \approx 0.90\,\frac{\mu_0 I}{R} \ >\ B_{circ}(0) = \frac{\mu_0 I}{2R} \]Integrating the square's axial field \(\displaystyle B_z = \dfrac{\mu_0 I R^2}{2\pi\,(z^2 + R^2/4)\sqrt{z^2 + R^2/2}}\):\[\mathfrak{I}_{sq}(L) = \frac{4\mu_0 I}{\pi}\arctan\frac{L}{\sqrt{L^2 + R^2/2}} \ \to\ \mu_0 I \]\[L = R:\quad \mathfrak{I}_{sq} = 0.87\,\mu_0 I \ >\ \mathfrak{I}_{circ} = 0.71\,\mu_0 I \]Answer: (a) \(\displaystyle d\mathfrak{I}/dL > 0\); (b), (c) \(\displaystyle \mathfrak{I}(\infty) = \mu_0 I\), with \(\displaystyle \mathfrak{I}(L) = \mu_0 I L/\sqrt{R^2 + L^2}\); (d) \(\displaystyle \mathfrak{I}_{sq}(L) > \mathfrak{I}_{circ}(L)\) at every finite \(\displaystyle L\), both \(\displaystyle \to \mu_0 I\).

  6. Exercise 28

    A multirange current meter can be constructed by using a galvanometer circuit as shown in Fig. 4.5. We want a current meter that can measure 10mA, 100mA and 1A using a galvanometer of resistance 10Ω\displaystyle 10 \Omega and that prduces maximum deflection for current of 1mA. Find S1,S2\displaystyle S_1, S_2 and S3\displaystyle S_3 that have to be used NCERT_Question_Class12_Physics_Exemplar_Ch4_Q28
    NCERT’s answer
    \(\displaystyle i_{\mathrm{G}} . G=\left(i_1-i_{\mathrm{G}}\right)\left(S_1+S_2+S_3\right) \quad\) for \(\displaystyle i_1=10 \mathrm{~mA}\) \(\displaystyle i_{\mathrm{G}}\left(G+S_1\right)=\left(i_2-i_{\mathrm{G}}\right)\left(S_2+S_3\right) \quad\) for \(\displaystyle i_2=100 \mathrm{~mA}\) and \(\displaystyle i_{\mathrm{G}}\left(G+S_1+S_2\right)=\left(i_3-i_{\mathrm{G}}\right)\left(S_3\right)\) for \(\displaystyle i_3=1 \mathrm{~A}\) gives \(\displaystyle S_1=1 \mathrm{~W}, S_2=0.1 \mathrm{~W}\) and \(\displaystyle S_3=0.01 \mathrm{~W}\)
    \(\displaystyle I_g = 1\ \mathrm{mA}\), \(\displaystyle G = 10\ \Omega\), so the galvanometer voltage is \(\displaystyle I_gG = 10\ \mathrm{mV}\). Caution: resistors between \(\displaystyle G\) and the chosen terminal join the galvanometer branch; only those beyond it form the shunt.NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q28$\displaystyle 10$ mA terminal: \(\displaystyle G\) in parallel with \(\displaystyle S_1+S_2+S_3\):\[I_g G = (10-1)\ \mathrm{mA}\,(S_1+S_2+S_3) \ \Rightarrow\ S_1+S_2+S_3 = \frac{10\ \mathrm{mV}}{9\ \mathrm{mA}} = \frac{10}{9}\ \Omega \]$\displaystyle 100$ mA terminal: \(\displaystyle G+S_1\) in parallel with \(\displaystyle S_2+S_3\):\[I_g\,(G+S_1) = 99\ \mathrm{mA}\,(S_2+S_3) \]\[1\,(10+S_1) = 99\left(\tfrac{10}{9}-S_1\right) \ \Rightarrow\ S_1 = 1\ \Omega, \quad S_2+S_3 = \tfrac{1}{9}\ \Omega \]$\displaystyle 1$ A terminal: \(\displaystyle G+S_1+S_2\) in parallel with \(\displaystyle S_3\):\[I_g\,(G+S_1+S_2) = 999\ \mathrm{mA}\;S_3 \]\[1\left(11+\tfrac{1}{9}-S_3\right) = 999\,S_3 \ \Rightarrow\ S_3 = \tfrac{1}{90}\ \Omega \approx 0.011\ \Omega, \quad S_2 = \tfrac{1}{9}-\tfrac{1}{90} = 0.1\ \Omega \]Answer: \(\displaystyle S_1 = 1\ \Omega\), \(\displaystyle S_2 = 0.1\ \Omega\), \(\displaystyle S_3 = \tfrac{1}{90}\ \Omega \approx 0.011\ \Omega\).

  7. Exercise 29

    Five long wires A, B, C, D and E, each carrying current I\displaystyle I are arranged to form edges of a pentagonal prism as shown in Fig. 4.6. Each carries current out of the plane of paper.
    (a)
    What will be magnetic induction at a point on the axis O? Axis is at a distance R\displaystyle R from each wire.
    (b)
    What will be the field if current in one of the wires (say A) is switched off?
    (c)
    What if current in one of the wire (say) A\displaystyle A is reversed?
    NCERT_Question_Class12_Physics_Exemplar_Ch4_Q29
    NCERT’s answer
    (a)
    zero
    (b)
    \(\displaystyle \frac{\mu_0}{2 \pi} \frac{i}{R}\) perpendicular to AO towards left.
    (c)
    \(\displaystyle \frac{\mu_0}{\pi} \frac{i}{R}\) perpendicular to AO towards left.
    Each wire gives, at \(\displaystyle O\), a field of equal size, tangential (perpendicular to its line to \(\displaystyle O\)):\[B_0 = \frac{\mu_0 I}{2\pi R} \](a) The five vectors of size \(\displaystyle B_0\) are \(\displaystyle 72^\circ\) apart; head to tail they close a regular pentagon.NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q29\[\mathbf{B}_A + \mathbf{B}_B + \mathbf{B}_C + \mathbf{B}_D + \mathbf{B}_E = \mathbf{0} \ \Rightarrow\ B_O = 0 \](b) A switched off:\[\mathbf{B}_O = \mathbf{B}_B + \mathbf{B}_C + \mathbf{B}_D + \mathbf{B}_E = -\mathbf{B}_A \ \Rightarrow\ B_O = \frac{\mu_0 I}{2\pi R} \](c) A reversed, \(\displaystyle \mathbf{B}_A \to -\mathbf{B}_A\):\[\mathbf{B}_O = (\mathbf{B}_B + \mathbf{B}_C + \mathbf{B}_D + \mathbf{B}_E) - \mathbf{B}_A = -2\mathbf{B}_A \ \Rightarrow\ B_O = \frac{\mu_0 I}{\pi R} \]At \(\displaystyle O\), \(\displaystyle \mathbf{B}_A\) points to the right (current out of the page, \(\displaystyle O\) below A), so (b) and (c) point left, along EB.Answer: (a) \(\displaystyle 0\); (b) \(\displaystyle \dfrac{\mu_0 I}{2\pi R}\); (c) \(\displaystyle \dfrac{\mu_0 I}{\pi R}\); in (b) and (c) opposite to \(\displaystyle \mathbf{B}_A\).