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NCERT Exemplar · Class 12 Physics Electromagnetic Induction

32 questions

Multiple Choice Questions (Type II) 7–10 (part 2 of 5)

  1. Exercise 7

    A metal plate is getting heated. It can be because
    (a)
    a direct current is passing through the plate.
    (b)
    it is placed in a time varying magnetic field.
    (c)
    it is placed in a space varying magnetic field, but does not vary with time.
    (d)
    a current (either direct or alternating) is passing through the plate.
    NCERT’s answer
    (a), (b), (d)
    (a), (b), (d) Heating needs a current in the plate, either sent through it or induced in it as eddy currents by a changing flux. A static field, however non-uniform, induces none.\[P = I^2R > 0 \quad (\text{DC or AC current in the plate}) \] \[\frac{dB}{dt}\neq 0 \;\Rightarrow\; \varepsilon = -\frac{d\Phi}{dt}\neq 0 \;\Rightarrow\; \text{eddy currents} \;\Rightarrow\; I^2R \text{ heating} \] \[\text{time-independent field: } \frac{d\Phi}{dt} = 0 \;\Rightarrow\; \text{no eddy current, no heating} \quad (\text{so (c) fails}) \]

  2. Exercise 8

    An e.m.f is produced in a coil, which is not connected to an external voltage source. This can be due to
    (a)
    the coil being in a time varying magnetic field.
    (b)
    the coil moving in a time varying magnetic field.
    (c)
    the coil moving in a constant magnetic field.
    (d)
    the coil is stationary in external spatially varying magnetic field, which does not change with time.
    NCERT’s answer
    (a), (b), (c)
    (a), (b), (c) An e.m.f. appears whenever the flux through the coil changes, by a changing field or by the coil's own motion. A stationary coil in a time-independent field sees constant flux.\[\varepsilon = -\frac{d\Phi}{dt}, \qquad \Phi = BA\cos\theta \] \[\text{(a) } \frac{\partial B}{\partial t}\neq 0 \Rightarrow \frac{d\Phi}{dt}\neq 0 \qquad \text{(b) } B,\ A\cos\theta \text{ both can change} \Rightarrow \frac{d\Phi}{dt}\neq 0 \] \[\text{(c) moving or turning in a non-uniform field, or entering/leaving it} \Rightarrow \Delta\Phi\neq 0 \] \[\text{(d) fixed coil, fixed field: } \frac{d\Phi}{dt} = 0 \Rightarrow \varepsilon = 0 \]

  3. Exercise 9

    The mutual inductance M12\displaystyle M_{12} of coil 1\displaystyle 1 with respect to coil 2\displaystyle 2
    (a)
    increases when they are brought nearer.
    (b)
    depends on the current passing through the coils.
    (c)
    increases when one of them is rotated about an axis.
    (d)
    is the same as M21\displaystyle \mathrm{M}_{21} of coil 2\displaystyle 2 with respect to coil 1.
    NCERT’s answer
    (a), (d)
    (a), (d) Mutual inductance is fixed by geometry alone: the nearer the coils, the more flux of one links the other. It does not depend on the currents, and rotation may raise or lower it.\[\Phi_1 = M_{12}\,I_2, \qquad \Phi_2 = M_{21}\,I_1 \] \[M_{12} = M_{21} \quad (\text{reciprocity}) \] \[\text{distance}\downarrow \Rightarrow \text{flux linkage}\uparrow \Rightarrow M\uparrow \] \[\text{rotation} \Rightarrow \text{relative orientation changes} \Rightarrow M \text{ may increase or decrease} \]

  4. Exercise 10

    A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because
    (a)
    the magnetic field is constant.
    (b)
    the magnetic field is in the same plane as the circular coil and it may or may not vary.
    (c)
    the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably.
    (d)
    there is a constant magnetic field in the perpendicular (to the plane of the coil) direction.
    NCERT’s answer
    (b), (c)
    (b), (c) Only the component of \(\displaystyle \mathbf{B}\) perpendicular to the coil's plane links the loop, and the area grows. No e.m.f. needs either no perpendicular component, or \(\displaystyle B_\perp\) falling just enough to keep the flux constant.\[\Phi = B_\perp A = B_\perp\,\pi r^2, \qquad \varepsilon = -\frac{d\Phi}{dt} \] \[\text{(b) } \mathbf{B}\ \text{in the plane} \Rightarrow B_\perp = 0 \Rightarrow \Phi = 0 \Rightarrow \varepsilon = 0 \] \[\text{(c) } B_\perp\,\pi r^2 = \text{const} \Rightarrow \frac{d\Phi}{dt} = 0 \Rightarrow \varepsilon = 0 \] \[\text{(d) } B_\perp = \text{const},\ r\uparrow \Rightarrow \Phi\uparrow \Rightarrow \varepsilon\neq 0 \]