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NCERT Exemplar · Class 12 Physics Magnetism and Matter

25 questions

Short Answer Questions 16–20 (part 4 of 5)

  1. Exercise 16

    Verify the Gauss's law for magnetic field of a point dipole of dipole moment m\displaystyle \mathbf{m} at the origin for the surface which is a sphere of radius R\displaystyle R.
    NCERT’s answer
    \(\displaystyle \mathbf{B}=\frac{\mu_0}{4 \pi} \frac{3 \mathbf{m} . \hat{\mathbf{r}}}{r^3}, m=m \hat{\mathbf{k}}\) \[\begin{aligned} & d \mathbf{s}=\hat{\mathbf{r}} . \mathbf{r}^2 \sin \theta d \theta d \\ & 0 \leq \theta \leq \pi, 0 \leq \phi \leq \end{aligned} \] \[\begin{aligned} \oint \mathbf{B} . d s & =\frac{\mu_0 m}{4 \pi} \int \frac{3 \cos \theta}{r^3} r^2 \sin \theta d \theta \\ & =0[\text { due to } \theta \text { integral }] . \end{aligned} \] NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q16_ncert
    Point dipole \(\displaystyle \mathbf{m}=m\hat{\mathbf{z}}\) at the origin; its field at \(\displaystyle (r,\theta)\): \[B_r=\frac{\mu_0}{4\pi}\,\frac{2m\cos\theta}{r^{3}},\qquad B_\theta=\frac{\mu_0}{4\pi}\,\frac{m\sin\theta}{r^{3}} \] NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q16 On the sphere \(\displaystyle r=R\), \(\displaystyle d\mathbf{S}=R^{2}\sin\theta\,d\theta\,d\phi\;\hat{\mathbf{r}}\); \(\displaystyle B_\theta\) is tangential, so only \(\displaystyle B_r\) gives flux: \[\Phi_B=\oint\mathbf{B}\cdot d\mathbf{S}=\frac{\mu_0}{4\pi}\,\frac{2m}{R}\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta\cos\theta\,d\theta=\frac{\mu_0 m}{R}\left[\frac{\sin^{2}\theta}{2}\right]_0^{\pi}=0 \] \[\Phi_{\text{upper}}=+\frac{\mu_0 m}{2R},\qquad \Phi_{\text{lower}}=-\frac{\mu_0 m}{2R}\ \Rightarrow\ \Phi_B=0\ \ \text{for every }R \] Answer: \(\displaystyle \oint\mathbf{B}\cdot d\mathbf{S}=0\) -- equal flux leaves the upper cap and enters the lower cap.

  2. Exercise 17

    Three identical bar magnets are rivetted together at centre in the same plane as shown in Fig. 5.1. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the Fig. 5.1. Determine the poles of the remaining two. NCERT_Question_Class12_Physics_Exemplar_Ch5_Q17
    NCERT’s answer
    Net \(\displaystyle m=0\). Only possibility is shown in Fig. NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q17_ncert
    No motion in a varying field means no torque and no force, whatever \(\displaystyle \mathbf{B}\) is: \[\boldsymbol{\tau}=\mathbf{m}_{net}\times\mathbf{B}=\mathbf{0}\ \ \text{for all }\mathbf{B}\ \Rightarrow\ \mathbf{m}_1+\mathbf{m}_2+\mathbf{m}_3=\mathbf{0} \] Identical magnets, \(\displaystyle |\mathbf{m}_i|=m\), so the three moments must be \(\displaystyle 120^\circ\) apart. With \(\displaystyle \mathbf{m}_1\) pointing up (\(\displaystyle 90^\circ\)): \[\mathbf{m}_1+\mathbf{m}_2+\mathbf{m}_3=m\left[(0,\,1)+\left(-\tfrac{\sqrt3}{2},\,-\tfrac12\right)+\left(\tfrac{\sqrt3}{2},\,-\tfrac12\right)\right]=\mathbf{0} \] NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q17 A moment points from S to N, so \(\displaystyle \mathbf{m}_2,\mathbf{m}_3\) (at \(\displaystyle 210^\circ,\,330^\circ\)) put N at the two lower ends. Answer: N at lower-left and lower-right; S at upper-left and upper-right.

  3. Exercise 18

    Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole p\displaystyle \mathbf{p} in an electrostatic field E\displaystyle \mathbf{E} and (ii) magnetic dipole m\displaystyle \mathbf{m} in a magnetic field B\displaystyle \mathbf{B}. Write down a set of conditions on E,B,p,m\displaystyle \mathbf{E}, \mathbf{B}, \mathbf{p}, \mathbf{m} so that the two motions are verified to be identical. (Assume identical initial conditions.)
    NCERT’s answer
    \(\displaystyle E(r)=c B(r), p=\frac{m}{c}\). Mass and moment of inertia of dipoles are equal.
    Torque and force take the same form for the two dipoles: \[\boldsymbol{\tau}=\mathbf{p}\times\mathbf{E},\qquad \mathbf{F}=\nabla(\mathbf{p}\cdot\mathbf{E}) \] \[\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B},\qquad \mathbf{F}=\nabla(\mathbf{m}\cdot\mathbf{B}) \] For rotation about the centre, moment of inertia \(\displaystyle I\), angle \(\displaystyle \theta\) to the field: \[I\ddot{\theta}=-pE\sin\theta,\qquad I\ddot{\theta}=-mB\sin\theta \] Both match when \[\mathbf{p}\parallel\mathbf{m},\qquad \mathbf{E}\parallel\mathbf{B},\qquad pE=mB\ \text{ at every point} \] with equal mass and \(\displaystyle I\). Units agree: \(\displaystyle \mathrm{C\,m\cdot N\,C^{-1}}=\mathrm{A\,m^{2}\cdot T}=\mathrm{N\,m}\). Answer: \(\displaystyle \mathbf{p}\parallel\mathbf{m}\), \(\displaystyle \mathbf{E}\parallel\mathbf{B}\), \(\displaystyle pE=mB\) everywhere, same mass and \(\displaystyle I\).

  4. Exercise 19

    A bar magnet of magnetic moment m\displaystyle m and moment of inertia I\displaystyle I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T\displaystyle T be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field B\displaystyle \mathbf{B}. What would be the similar period T′\displaystyle T^{\prime} for each piece?
    NCERT’s answer
    \(\displaystyle T=2 \pi \sqrt{\frac{I}{m B}} \quad I^{\prime}=\frac{1}{2} \times \frac{1}{4} I\) and \(\displaystyle m^{\prime}=\frac{m}{2} . \quad T^{\prime}=\frac{1}{2} T\)
    Original magnet: \[T=2\pi\sqrt{\frac{I}{mB}} \] The cut halves the length \(\displaystyle L\) and mass \(\displaystyle M\) but leaves the pole strength \(\displaystyle q_m\) unchanged, so \(\displaystyle m\) halves: \[m=q_mL\ \Rightarrow\ m'=q_m\frac{L}{2}=\frac{m}{2} \] \[I=\frac{ML^{2}}{12}\ \Rightarrow\ I'=\frac{(M/2)(L/2)^{2}}{12}=\frac{I}{8} \] \[T'=2\pi\sqrt{\frac{I/8}{(m/2)\,B}}=2\pi\sqrt{\frac{I}{4mB}}=\frac{T}{2} \] Answer: \(\displaystyle T'=T/2\).

  5. Exercise 20

    Use (i) the Ampere's law for H\displaystyle \mathbf{H} and (ii) continuity of lines of B\displaystyle \mathbf{B}, to conclude that inside a bar magnet, (a) lines of H\displaystyle \mathbf{H} run from the N\displaystyle N pole to S\displaystyle S pole, while (b) lines of B\displaystyle \mathbf{B} must run from the S\displaystyle S pole to N\displaystyle N pole.
    NCERT’s answer
    Consider a line of \(\displaystyle \mathbf{B}\) through the bar magnet. It must be closed. Let C be the amperian loop. \[\begin{aligned} & \int_Q^P \mathbf{H} . d \mathbf{l}=\int_Q^P \frac{\mathbf{B}}{\mu_0} . d \boldsymbol{l}>0 \\ & \oint_{P Q P} \mathbf{H} . d \mathbf{l}=0 \end{aligned} \] \[\int_p^Q \mathbf{H} . d \mathbf{l}<0 \] P → Q is inside the bar. Hence \(\displaystyle \mathbf{H}\) is making an obtuse angle with \(\displaystyle d \boldsymbol{l}\). NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q20_ncert
    A magnet carries no free current, so for any closed loop \(\displaystyle \Gamma\): \[\oint_\Gamma\mathbf{H}\cdot d\mathbf{l}=I_{free}=0 \] NCERT_Solution_Class12_Physics_Exemplar_Ch5_Q20 Take \(\displaystyle \Gamma\): outside from \(\displaystyle N\) to \(\displaystyle S\), then back through the magnet from \(\displaystyle S\) to \(\displaystyle N\). Outside, \(\displaystyle \mathbf{H}=\mathbf{B}/\mu_0\) runs \(\displaystyle N\to S\): \[\int_{out}\mathbf{H}\cdot d\mathbf{l}>0\ \Rightarrow\ \int_{in}\mathbf{H}\cdot d\mathbf{l}=-\int_{out}\mathbf{H}\cdot d\mathbf{l}<0 \] Inside, \(\displaystyle \mathbf{H}\) opposes the \(\displaystyle S\to N\) path: (a) \(\displaystyle N\to S\). \[\nabla\cdot\mathbf{B}=0\ \Rightarrow\ \text{lines of }\mathbf{B}\text{ close on themselves} \] Outside they run \(\displaystyle N\to S\), so inside they must return \(\displaystyle S\to N\): (b). Answer: inside the magnet, \(\displaystyle \mathbf{H}\) runs \(\displaystyle N\to S\) and \(\displaystyle \mathbf{B}\) runs \(\displaystyle S\to N\).