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NCERT Exemplar · Class 12 Physics Magnetism and Matter

25 questions

Long Answer Questions 21–25 (part 5 of 5)

  1. Exercise 21

    Verify the Ampere's law for magnetic field of a point dipole of dipole moment m=mk^\displaystyle \mathbf{m}=m \hat{\mathbf{k}}. Take C as the closed curve running clockwise along (i) the z\displaystyle z-axis from z=a>0\displaystyle z=a>0 to z=R\displaystyle z=R; (ii) along the quarter circle of radius R\displaystyle R and centre at the origin, in the first quadrant of x−z\displaystyle x-z plane; (iii) along the x\displaystyle x-axis from x=R\displaystyle x=R to x=a\displaystyle x=a, and (iv) along the quarter circle of radius a\displaystyle a and centre at the origin in the first quadrant of x−z\displaystyle x-z plane.
    NCERT’s answer
    (i)
    Along \(\displaystyle z\) axis
    \[\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{2 \mathbf{m}}{r^3} \]
    \[\int_a^R \mathbf{B} \cdot d \mathbf{l}=\frac{\mu_0}{4 \pi} 2 m \int_a^R \frac{d z}{z^3}=\frac{\mu_0 m}{2 \pi}\left(-\frac{1}{2}\right)\left(\frac{1}{R^2}-\frac{1}{a^2}\right) \]
    (ii)
    Along the quarter circle of radius \(\displaystyle R\)
    \[B_{\emptyset}=\frac{\mu_0}{4 \pi} \frac{-\mathbf{m} \cdot \hat{\boldsymbol{\theta}}}{R^3}=\frac{-\mu_0}{4 \pi} \frac{m}{R^3}(-\sin \theta) \]
    \[\begin{aligned} & \mathbf{B} \cdot d \boldsymbol{l}=\frac{\mu_0 m}{4 \pi R^2} \sin \theta d \\ & \int_0^{\frac{\pi}{2}} \overrightarrow{\mathrm{~B}} \cdot \overrightarrow{d l}=\frac{\mu_0 m}{4 \pi R^2} \end{aligned} \]
    (iii)
    Along \(\displaystyle x\)-axis
    \[\begin{aligned} & \mathbf{B}=\frac{\mu_0}{4 \pi}\left(\frac{-m}{x^3}\right) \\ & \int \mathbf{B} \cdot d \mathbf{l}=0 \end{aligned} \]
    (iv)
    Along the quarter circle of radius \(\displaystyle a\)
    \[\mathbf{B} \cdot d \mathbf{l}=\frac{-\mu_0 m}{4 \pi a^2} \sin \theta d \theta, \int \mathbf{B} \cdot d \mathbf{l}=-\frac{-\mu_0 m}{4 \pi a^2} \int_0^{\frac{\pi}{2}} \sin \theta d \theta=\frac{-\mu_0 m}{4 \pi a^2} \]
    Add \(\displaystyle \oint_C \mathbf{B} \cdot d \mathbf{l}=0\)
    NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q21_ncert
    Dipole field at \(\displaystyle (r,\theta)\), with \(\displaystyle \theta\) measured from the \(\displaystyle z\)-axis: \[B_r=\frac{\mu_0}{4\pi}\frac{2m\cos\theta}{r^3},\qquad B_\theta=\frac{\mu_0}{4\pi}\frac{m\sin\theta}{r^3} \] Path: \(\displaystyle A(z=a)\to B(z=R)\to C(x=R)\to D(x=a)\to A\). NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q21 Along \(\displaystyle AB\) (\(\displaystyle \theta=0\), \(\displaystyle d\mathbf{l}=dr\,\hat{\mathbf r}\)): \[\int_A^B\mathbf{B}\cdot d\mathbf{l}=\int_a^R\frac{\mu_0}{4\pi}\frac{2m}{r^3}\,dr=\frac{\mu_0 m}{4\pi}\left(\frac1{a^2}-\frac1{R^2}\right) \] Along arc \(\displaystyle BC\) (\(\displaystyle r=R\), \(\displaystyle d\mathbf{l}=R\,d\theta\,\hat{\boldsymbol\theta}\), \(\displaystyle \theta:0\to\pi/2\)): \[\int_B^C\mathbf{B}\cdot d\mathbf{l}=\int_0^{\pi/2}\frac{\mu_0}{4\pi}\frac{m\sin\theta}{R^3}\,R\,d\theta=\frac{\mu_0 m}{4\pi R^2} \] Along \(\displaystyle CD\) (\(\displaystyle \theta=\pi/2\)): \(\displaystyle B_r=0\) and \(\displaystyle B_\theta\perp d\mathbf{l}\), so \[\int_C^D\mathbf{B}\cdot d\mathbf{l}=0 \] Along arc \(\displaystyle DA\) (\(\displaystyle r=a\), \(\displaystyle \theta:\pi/2\to0\)): \[\int_D^A\mathbf{B}\cdot d\mathbf{l}=\int_{\pi/2}^{0}\frac{\mu_0}{4\pi}\frac{m\sin\theta}{a^3}\,a\,d\theta=-\frac{\mu_0 m}{4\pi a^2} \] Sum: \[\oint\mathbf{B}\cdot d\mathbf{l}=\frac{\mu_0 m}{4\pi}\left[\frac1{a^2}-\frac1{R^2}+\frac1{R^2}+0-\frac1{a^2}\right]=0 \] No current threads the loop, so \(\displaystyle I_{enc}=0\): \[\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{enc}=0 \] Answer: \(\displaystyle \oint\mathbf{B}\cdot d\mathbf{l}=0=\mu_0 I_{enc}\); Ampere's law holds for the dipole field.

  2. Exercise 22

    What are the dimensions of χ\displaystyle \chi, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ\displaystyle \chi, upto a constant by constructing a quantity of dimensions of χ\displaystyle \chi, out of parameters of the atom: e,m,v,R\displaystyle e, m, v, R and μ0\displaystyle \mu_0. Here, m\displaystyle m is the electronic mass, v\displaystyle v is electronic velocity, R\displaystyle R is Bohr radius. Estimate the number so obtained and compare with the value of ∣χ∣∼10−5\displaystyle |\chi| \sim 10^{-5} for many solid materials.
    NCERT’s answer
    \(\displaystyle \chi\) is dimensionless. \(\displaystyle \chi\) depends on magnetic moment induced when \(\displaystyle H\) is turned on. \(\displaystyle H\) couples to atomic electrons through its charge \(\displaystyle e\). The effect on \(\displaystyle m\) is via current \(\displaystyle I\) which involves another factor of ‘\(\displaystyle e\)’. The combination " \(\displaystyle \mu_0 e^2\) " does not depend on the “charge” Q dimension. \[\chi=\mu_0 e^2 m^\alpha v^\beta R^\gamma \] \[\mu_0 c^2=\frac{1}{c^2} \frac{e^2}{\varepsilon_0} \sim \frac{1}{c^2} \frac{e^2}{\varepsilon_0 R} \cdot R \sim \frac{\text { Energy length }}{c^2} \] \[\begin{aligned} & {[\chi]=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{Q}^0=\frac{\mathrm{ML}^3 \mathrm{~T}^{-2}}{L^2 T^{-2}} M^\alpha\left(\frac{\mathrm{L}}{\mathrm{~T}}\right)^\beta \mathrm{L}^\gamma \mathrm{Q}^0} \\ & \alpha=-1, \beta=0, \gamma=-1 \\ & \chi=\frac{\mu_0 e^2}{m R} \sim \frac{10^{-6} \times 10^{-38}}{10^{-30} \times 10^{-10}} \sim 10^{-4} . \end{aligned} \]
    \[\chi=\frac{M}{H},\qquad [M]=[H]=\mathrm{A\,m^{-1}}\ \Rightarrow\ [\chi]=1 \] So \(\displaystyle \chi\) is dimensionless. Try \(\displaystyle \chi\sim\mu_0^{a}e^{b}m^{c}v^{d}R^{f}\), with \[[\mu_0]=\mathrm{kg\,m\,s^{-2}A^{-2}},\quad [e]=\mathrm{A\,s},\quad [m]=\mathrm{kg},\quad [v]=\mathrm{m\,s^{-1}},\quad [R]=\mathrm{m} \] The powers of kg, m, s, A must each vanish: \[a+c=0,\qquad a+d+f=0,\qquad -2a+b-d=0,\qquad -2a+b=0 \] \[\Rightarrow\ b=2a,\quad d=0,\quad c=-a,\quad f=-a \] Taking \(\displaystyle a=1\) (the speed \(\displaystyle v\) drops out): \[\chi\sim\frac{\mu_0e^2}{mR} \] With \(\displaystyle R=0.53\times10^{-10}\,\mathrm{m}\): \[\chi\sim\frac{(4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(1.6\times10^{-19}\,\mathrm{C})^2}{(9.1\times10^{-31}\,\mathrm{kg})(0.53\times10^{-10}\,\mathrm{m})}\approx7\times10^{-4} \] \[\frac{7\times10^{-4}}{10^{-5}}\approx70 \] Both are \(\displaystyle \ll1\): the same tiny scale, within two orders of magnitude. Answer: \(\displaystyle \chi\) is dimensionless; \(\displaystyle \chi\sim\mu_0e^2/(mR)\approx7\times10^{-4}\), about \(\displaystyle 70\times\) the solid-state \(\displaystyle |\chi|\sim10^{-5}\).

  3. Exercise 23

    Assume the dipole model for earth's magnetic field B\displaystyle B which is given by BV=\displaystyle B_V= vertical component of magnetic field =μ04π2mcos⁡θr3\displaystyle =\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^3} BH=\displaystyle B_H= Horizontal component of magnetic field =μ04πsin⁡θmr3\displaystyle =\frac{\mu_0}{4 \pi} \frac{\sin \theta m}{r^3} θ=90∘−\displaystyle \theta=90^{\circ}- lattitude as measured from magnetic equator. Find loci of points for which (i) ∣B∣\displaystyle |\mathbf{B}| is minimum; (ii) dip angle is zero; and (iii) dip angle is ±45∘\displaystyle \pm 45^{\circ}.
    NCERT’s answer
    (i)
    \(\displaystyle |\mathbf{B}|=\frac{\mu_0}{4 \pi} \frac{m}{R^3}\left(4 \cos ^2 \theta+\sin ^2 \theta\right)^{1 / 2}\)
    \(\displaystyle \frac{|\mathbf{B}|^2}{\left(\dfrac{\mu_0}{4 \pi R^3}\right)^2 m^2}=3 \cos ^2 \theta+1\), minimum at \(\displaystyle \theta=\frac{\pi}{2}\).
    \(\displaystyle |\mathbf{B}|\) is minimum at magnetic equator.
    (ii)
    \(\displaystyle \tan\) (dip angle) \(\displaystyle =\frac{B_V}{B_H}=2 \cot\)
    at \(\displaystyle \theta=\frac{\pi}{2}\) dip angle vanishes. Magnetic equator is again the locus.
    (iii)
    Dip angle is ± $\displaystyle 45$° when \(\displaystyle \left|\frac{B_V}{B_H}\right|=1\)
    \[2 \cot \theta=1 \]
    \(\displaystyle \theta=\tan ^{-1} 2\) is the locus.
    \[|\mathbf B|=\sqrt{B_V^2+B_H^2}=\frac{\mu_0 m}{4\pi r^3}\sqrt{4\cos^2\theta+\sin^2\theta}=\frac{\mu_0 m}{4\pi r^3}\sqrt{1+3\cos^2\theta} \] (i) At a given \(\displaystyle r\), \(\displaystyle |\mathbf B|\) is least when \(\displaystyle \cos\theta=0\): \[\theta=90^\circ,\qquad |\mathbf B|_{min}=\frac{\mu_0 m}{4\pi r^3} \] Locus: the magnetic equator.(ii) Dip angle \(\displaystyle \delta\): \[\tan\delta=\frac{B_V}{B_H}=\frac{2\cos\theta}{\sin\theta}=2\cot\theta \] \[\delta=0\ \Rightarrow\ \cot\theta=0\ \Rightarrow\ \theta=90^\circ \] Again the magnetic equator.(iii) \[\delta=\pm45^\circ\ \Rightarrow\ 2\cot\theta=\pm1\ \Rightarrow\ \tan\theta=\pm2 \] \[\theta=63.4^\circ\ \text{or}\ 116.6^\circ,\qquad \lambda=90^\circ-\theta=\pm26.6^\circ \] NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q23 Answer: (i) and (ii) \(\displaystyle \theta=90^\circ\), the magnetic equator; (iii) \(\displaystyle \theta=63.4^\circ\) and \(\displaystyle 116.6^\circ\), the circles of magnetic latitude \(\displaystyle \pm26.6^\circ\).

  4. Exercise 24

    Consider the plane S\displaystyle S formed by the dipole axis and the axis of earth. Let P\displaystyle P be point on the magnetic equator and in S\displaystyle S. Let Q\displaystyle Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P\displaystyle P and Q\displaystyle Q.
    NCERT’s answer
    Refer to the adjacent Fig. 1. P is in S (needle will point both north) Declination \(\displaystyle =0\) P is also on magnetic equator. \[\therefore \operatorname{dip}=0 \] 2. Q is on magnetic equator. \[\therefore \operatorname{dip}=0 \] but declination \(\displaystyle =11.3^{\circ}\). NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q24_ncert
    Declination \(\displaystyle D\), dip \(\displaystyle I\). Plane \(\displaystyle S\) holds the earth's axis and the dipole axis, inclined at \(\displaystyle \alpha\approx11.3^\circ\). NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q24 At \(\displaystyle P\) (magnetic equator, in \(\displaystyle S\)): the geographic and the magnetic meridian planes through \(\displaystyle P\) both coincide with \(\displaystyle S\), and the magnetic equator gives zero dip (\(\displaystyle \mathbf{B}\) horizontal): \[D_P=0^\circ,\qquad I_P=0^\circ \] At \(\displaystyle Q\) (on both equators): \(\displaystyle OQ\perp S\), so the horizontal plane at \(\displaystyle Q\) is parallel to \(\displaystyle S\). On the magnetic equator \(\displaystyle \mathbf{B}\) is horizontal and parallel to the dipole axis; geographic north is parallel to the earth's axis: \[I_Q=0^\circ \] \[D_Q=\angle(\text{dipole axis},\ \text{earth's axis})=\alpha\approx11.3^\circ \] Answer: \(\displaystyle P\): declination \(\displaystyle D=0^\circ\), dip \(\displaystyle I=0^\circ\). \(\displaystyle Q\): declination \(\displaystyle D=\alpha\approx11.3^\circ\), dip \(\displaystyle I=0^\circ\).

  5. Exercise 25

    There are two current carrying planar coils made each from identical wires of length L\displaystyle L. C1\displaystyle \mathrm{C}_1 is circular (radius R\displaystyle R ) and C2\displaystyle \mathrm{C}_2 is square (side a\displaystyle a ). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform B\displaystyle \mathbf{B} and carry the same current. Find a\displaystyle a in terms of R\displaystyle R.
    Same wire: length \(\displaystyle L\), mass \(\displaystyle M\), wound into as many turns as it allows: \[n_1=\frac{L}{2\pi R},\qquad n_2=\frac{L}{4a} \] Moments, same current \(\displaystyle I\): \[m_1=n_1I\pi R^2=\frac{ILR}{2},\qquad m_2=n_2Ia^2=\frac{ILa}{4} \] Rotational inertia about an in-plane axis through the centre: \[J_1=\tfrac12MR^2\ \text{(ring)},\qquad J_2=\tfrac16Ma^2\ \text{(square frame)} \] Caution: a wire frame is \(\displaystyle Ma^2/6\), not the plate's \(\displaystyle Ma^2/12\). The printed \(\displaystyle a=3\pi R/4\) does not follow from correct working. Small oscillations in \(\displaystyle \mathbf{B}\): \(\displaystyle \nu=\dfrac1{2\pi}\sqrt{mB/J}\), so equal \(\displaystyle \nu\) needs \(\displaystyle m_1/J_1=m_2/J_2\): \[\frac{ILR/2}{MR^2/2}=\frac{ILa/4}{Ma^2/6}\ \Rightarrow\ \frac{IL}{MR}=\frac{3IL}{2Ma} \] \[a=\frac{3R}{2} \] Answer: \(\displaystyle a=\dfrac{3R}{2}\).