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NCERT Exemplar · Class 12 Physics Moving Charges and Magnetism

29 questions

Multiple Choice Questions (Type II) 7–11 (part 2 of 5)

  1. Exercise 7

    The gyro-magnetic ratio of an electron in an H-atom, according to Bohr model, is
    (a)
    independent of which orbit it is in.
    (b)
    negative.
    (c)
    positive.
    (d)
    increases with the quantum number n\displaystyle n.
    NCERT’s answer
    (a), (b)
    (a), (b) The electron's charge is negative, so \(\displaystyle \boldsymbol{\mu}\) opposes \(\displaystyle \mathbf{L}\); \(\displaystyle n\) cancels in the ratio.\[I=\frac{e}{T}=\frac{ev}{2\pi r},\qquad \mu=IA=\frac{ev}{2\pi r}\,\pi r^{2}=\frac{evr}{2} \]\[L=mvr\ \Rightarrow\ \frac{\mu}{L}=\frac{e}{2m},\qquad \boldsymbol{\mu}=-\frac{e}{2m}\,\mathbf{L} \]\[L=n\hbar\ \Rightarrow\ \mu_n=-n\,\frac{e\hbar}{2m}\quad(\text{grows with } n),\ \text{but}\ \gamma=\frac{\mu_n}{L}=-\frac{e}{2m} \]\[\gamma=-\frac{1.6\times10^{-19}\ \mathrm{C}}{2\times9.1\times10^{-31}\ \mathrm{kg}}\approx-8.8\times10^{10}\ \mathrm{C\,kg^{-1}} \]Answer: (a), (b)

  2. Exercise 8

    Consider a wire carrying a steady current, I\displaystyle I placed in a uniform magnetic field B\displaystyle \mathbf{B} perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,
    (a)
    motion of charges inside the conductor is unaffected by B\displaystyle \mathbf{B} since they do not absorb energy.
    (b)
    some charges inside the wire move to the surface as a result of B\displaystyle \mathbf{B}.
    (c)
    if the wire moves under the influence of B\displaystyle \mathbf{B}, no work is done by the force.
    (d)
    if the wire moves under the influence of B\displaystyle \mathbf{B}, no work is done by the magnetic force on the ions, assumed fixed within the wire.
    NCERT’s answer
    (b), (d)
    (b), (d) \(\displaystyle \mathbf{B}\) pushes the carriers sideways, so charge collects on one face. Each ion's magnetic force is perpendicular to its velocity, so it does no work. (a), (c) fail.\[\mathbf{F}_B=q\,\mathbf{v}\times\mathbf{B}\ \Rightarrow\ \mathbf{F}_B\cdot\mathbf{v}=0 \]NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q8\[qE_H=q\,v_dB\ \Rightarrow\ E_H=v_dB \]\[\text{ion moving with the wire at } \mathbf{u}:\ \mathbf{F}_B=q\,\mathbf{u}\times\mathbf{B}\ \Rightarrow\ \mathbf{F}_B\cdot\mathbf{u}=0\ \Rightarrow\ W=0 \]

  3. Exercise 9

    Two identical current carrying coaxial loops, carry current I\displaystyle I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C\displaystyle C,
    (a)
    ∮cB.dl=∓2μ0I\displaystyle \oint_c \mathbf{B} . \mathbf{d l}=\mp 2 \mu_0 I.
    (b)
    the value of ∮cB.dl\displaystyle \oint_c \mathbf{B} . \mathbf{d l} is independent of sense of C.
    (c)
    there may be a point on C where B\displaystyle \mathbf{B} and dl\displaystyle \mathbf{d l} are perpendicular.
    (d)
    B\displaystyle \mathbf{B} vanishes everywhere on C.
    NCERT’s answer
    (b), (c)
    (b), (c) The enclosed currents cancel, so the circulation is zero for either sense of C. B is not zero all along C, so \(\displaystyle \mathbf{B}\cdot d\mathbf{l}\) changes sign and vanishes somewhere: B is perpendicular to \(\displaystyle d\mathbf{l}\) there.NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q9\[I_{enc}=I+(-I)=0 \]\[\oint_C\mathbf{B}\cdot d\mathbf{l}=\mu_0I_{enc}=0\quad(\text{either sense of } C) \]\[\oint_C\mathbf{B}\cdot d\mathbf{l}=0,\ \ \mathbf{B}\neq\mathbf{0}\ \Rightarrow\ B\cos\theta\ \text{changes sign}\ \Rightarrow\ \theta=90^\circ\ \text{at some point} \]Answer: (b), (c)

  4. Exercise 10

    A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v\displaystyle \mathbf{v} and a positron enters via opposite face with velocity - v\displaystyle \mathbf{v}. At this instant,
    (a)
    the electric forces on both the particles cause identical accelerations.
    (b)
    the magnetic forces on both the particles cause equal accelerations.
    (c)
    both particles gain or loose energy at the same rate.
    (d)
    the motion of the centre of mass (CM) is determined by B\displaystyle \mathbf{B} alone.
    NCERT’s answer
    (b), (c), (d)
    (b), (c), (d) The magnetic forces are identical and the electric power is the same for both; the electric forces cancel for the CM. (a) fails: the accelerations are opposite.\[\mathbf{F}_E^{-}=-e\mathbf{E},\quad \mathbf{F}_E^{+}=+e\mathbf{E}\ \Rightarrow\ \mathbf{a}_-=-\mathbf{a}_+ \]\[\mathbf{F}_B^{-}=(-e)\,\mathbf{v}\times\mathbf{B},\qquad \mathbf{F}_B^{+}=(+e)(-\mathbf{v})\times\mathbf{B}=-e\,\mathbf{v}\times\mathbf{B} \]\[P^{-}=(-e\mathbf{E})\cdot\mathbf{v}=-e\,\mathbf{E}\cdot\mathbf{v},\qquad P^{+}=(e\mathbf{E})\cdot(-\mathbf{v})=-e\,\mathbf{E}\cdot\mathbf{v} \]\[\mathbf{F}_{net}=(-e\mathbf{E}+e\mathbf{E})-2e\,\mathbf{v}\times\mathbf{B}\ \Rightarrow\ \mathbf{a}_{CM}=\frac{\mathbf{F}_{net}}{2m}=-\frac{e}{m}\,\mathbf{v}\times\mathbf{B} \]Answer: (b), (c), (d)

  5. Exercise 11

    A charged particle would continue to move with a constant velocity in a region wherein,
    (a)
    E=0,B≠0\displaystyle \mathbf{E}=0, \mathbf{B} \neq 0.
    (b)
    E≠0,B≠0\displaystyle \mathbf{E} \neq 0, \mathbf{B} \neq 0.
    (c)
    E≠0,B=0\displaystyle \mathbf{E} \neq 0, \mathbf{B}=0.
    (d)
    E=0,B=0\displaystyle \mathbf{E}=0, \mathbf{B}=0.
    NCERT’s answer
    (a), (b), (d)
    (a), (b), (d) Constant velocity needs zero net force. (a) and (b) work only for the right orientation; (c) always accelerates the particle.\[\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})=\mathbf{0} \]\[\text{(a)}\ \ \mathbf{E}=0:\ \ q\,\mathbf{v}\times\mathbf{B}=\mathbf{0}\ \ \text{if}\ \ \mathbf{v}\parallel\mathbf{B} \]\[\text{(b)}\ \ \mathbf{E}=-\mathbf{v}\times\mathbf{B}\ \Rightarrow\ \mathbf{F}=\mathbf{0},\quad E=vB \]\[\text{(c)}\ \ \mathbf{B}=0:\ \ \mathbf{F}=q\mathbf{E}\neq\mathbf{0}\ \Rightarrow\ \mathbf{a}\neq\mathbf{0} \]\[\text{(d)}\ \ \mathbf{E}=\mathbf{B}=0:\ \ \mathbf{F}=\mathbf{0} \]Answer: (a), (b), (d)