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NCERT Exemplar · Class 12 Physics Electromagnetic Induction

32 questions

Short Answer Questions 18–21 (part 4 of 5)

  1. Exercise 18

    Consider a closed loop C in a magnetic field (Fig 6.7\displaystyle 6.7). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula ϕ=B1.dA1+B2.dA2+…\displaystyle \phi=\mathbf{B}_1 . \boldsymbol{d} \mathbf{A}_1+\boldsymbol{B}_2 . \boldsymbol{d} \mathbf{A}_2+\ldots. Now if we chose two different surfaces S1\displaystyle \mathrm{S}_1 and S2\displaystyle \mathrm{S}_2 having C as their edge, would we get the same answer for flux. Jusity your answer. NCERT_Question_Class12_Physics_Exemplar_Ch6_Q18
    NCERT’s answer
    One gets the same answer for flux. Flux can be throught of as the number of magnetic field lines passing through the surface (we draw \(\displaystyle \mathrm{d} N=B \Delta A\) lines in an area \(\displaystyle \Delta \mathrm{A} \perp\) to \(\displaystyle \mathbf{B}\)), As lines of of \(\displaystyle \mathbf{B}\) cannot end or start in space (they form closed loops) number of lines passing through surface \(\displaystyle S_1\) must be the same as the number of lines passing through the surface \(\displaystyle S_2\). NCERT_Solution_Class12_Physics_Exemplar_Ch6_Q18_ncert
    Yes, the same. S\(\displaystyle _1\) and S\(\displaystyle _2\) together enclose a volume. The outward normal on S\(\displaystyle _1\) is opposite to the normal chosen for the loop's flux, so S\(\displaystyle _1\) contributes \(\displaystyle -\phi_1\).\[\oint_{S_1+S_2}\mathbf{B}\cdot d\mathbf{A}=0 \quad (\text{no magnetic monopole}) \]\[\phi_2+(-\phi_1)=0 \]\[\phi_1=\phi_2 \]Answer: Yes, the flux is the same for S\(\displaystyle _1\) and S\(\displaystyle _2\), because the net flux of \(\displaystyle \mathbf{B}\) through the closed surface S\(\displaystyle _1+\)S\(\displaystyle _2\) is zero.

  2. Exercise 19

    Find the current in the wire for the configuration shown in Fig 6.8. Wire PQ has negligible resistance. B\displaystyle \mathbf{B}, the magnetic field is coming out of the paper. θ\displaystyle \theta is a fixed angle made by PQ travelling smoothly over two conducting parallel wires seperated by a distance d\displaystyle d. NCERT_Question_Class12_Physics_Exemplar_Ch6_Q19
    NCERT’s answer
    Motional electric field \(\displaystyle E\) along the dotted line CD (\(\displaystyle \perp\) to both \(\displaystyle \mathbf{v}\) and \(\displaystyle \mathbf{B}\) and along \(\displaystyle \mathbf{v} \times \mathbf{B}\)) \(\displaystyle =v B\) E.M.F. along PQ = (length PQ)×(Field along PQ) \[=\frac{d}{\cos \theta} \times v B \cos \theta=d v B . \] Therefore, \[I=\frac{d v B}{R} \text { and is independent of } q \text {. } \] NCERT_Solution_Class12_Physics_Exemplar_Ch6_Q19_ncert
    Only the component of \(\displaystyle v\) perpendicular to the rod cuts the field, and the rod's length between the rails is not \(\displaystyle d\). The \(\displaystyle \theta\) cancels.\[l=\frac{d}{\sin\theta},\qquad v_\perp=v\sin\theta \]\[\varepsilon=B\,l\,v_\perp=B\,\frac{d}{\sin\theta}\,(v\sin\theta)=Bvd \]\[I=\frac{\varepsilon}{R}=\frac{Bvd}{R} \]NCERT_Solution_Class12_Physics_Exemplar_Ch6_Q19Direction (Lenz): \(\displaystyle \phi_{\text{out}}\uparrow \Rightarrow\) induced field into the page \(\displaystyle \Rightarrow\) clockwise, so \(\displaystyle P\to Q\) in the rod and upward through \(\displaystyle R\).Answer: \(\displaystyle I=\dfrac{Bvd}{R}\), independent of \(\displaystyle \theta\), clockwise.

  3. Exercise 20

    A (current vs time) graph of the current passing through a solenoid is shown in Fig 6.9. For which time is the back electromotive force (u)\displaystyle (u) a maximum. If the back emf at t=3 s\displaystyle t=3 \mathrm{~s} is e\displaystyle e, find the back emf at t=7 s,15 s\displaystyle t=7 \mathrm{~s}, 15 \mathrm{~s} and 40s. OA, AB and BC are straight line segments. NCERT_Question_Class12_Physics_Exemplar_Ch6_Q20
    NCERT’s answer
    Maximum rate of change of current is in AB. So maximum back emf will be obtained between \(\displaystyle 5 \mathrm{~s}<t<10 \mathrm{~s}\). If \(\displaystyle u=L 1 / 5\) (for \(\displaystyle t=3 \mathrm{~s}, \frac{d I}{d t}=1 / 5\)) (\(\displaystyle L\) is a constant) For 5s \(\displaystyle <t<10 \mathrm{~s} \quad u_1=-L \frac{3}{5}=-\frac{3}{5} L=-3 e\) Thus at \(\displaystyle t=7 \mathrm{~s}, u_1=-3 e\). For \(\displaystyle 10 \mathrm{~s}<t<30 \mathrm{~s}\) \[u_2=L \frac{2}{20}=\frac{L}{10}=\frac{1}{2} e \] For \(\displaystyle t>30 \mathrm{~s} \quad u_2=0\)
    Back emf is largest where the graph is steepest. Current is zero after C.\[\varepsilon=-L\frac{dI}{dt} \]OA ($\displaystyle 0$ to $\displaystyle 5$ s): \[\frac{dI}{dt}=\frac{1-0}{5-0}=0.2\ \mathrm{A\,s^{-1}}\ \Rightarrow\ e=-0.2L \]AB ($\displaystyle 5$ s to $\displaystyle 10$ s): \[\frac{dI}{dt}=\frac{-2-1}{10-5}=-0.6\ \mathrm{A\,s^{-1}}\ \Rightarrow\ \varepsilon_{7}=+0.6L=-3e \]BC ($\displaystyle 10$ s to $\displaystyle 30$ s): \[\frac{dI}{dt}=\frac{0-(-2)}{30-10}=0.1\ \mathrm{A\,s^{-1}}\ \Rightarrow\ \varepsilon_{15}=-0.1L=\frac{e}{2} \]After C: \[\frac{dI}{dt}=0\ \Rightarrow\ \varepsilon_{40}=0 \]NCERT_Solution_Class12_Physics_Exemplar_Ch6_Q20Answer: Maximum between \(\displaystyle t=5\ \mathrm{s}\) and \(\displaystyle 10\ \mathrm{s}\) (segment AB); \(\displaystyle \varepsilon_{7}=-3e\), \(\displaystyle \varepsilon_{15}=e/2\), \(\displaystyle \varepsilon_{40}=0\).

  4. Exercise 21

    There are two coils A and B seperated by some distance. If a current of 2\displaystyle 2 A flows through A, a magnetic flux of 10−2 Wb\displaystyle 10^{-2} \mathrm{~Wb} passes through B (no current through B). If no current passes through A and a current of 1\displaystyle 1 A passes through B, what is the flux through A?
    NCERT’s answer
    Mutual inductance \(\displaystyle =\frac{10^{-2}}{2}=5 \mathrm{mH}\) \[\text { Flux }=5 \times 10^{-3} \times 1=5 \times 10^{-3} \mathrm{~Wb} . \]
    Mutual inductance is the same both ways.\[\phi_B=M\,I_A\ \Rightarrow\ M=\frac{10^{-2}\ \mathrm{Wb}}{2\ \mathrm{A}}=5\times10^{-3}\ \mathrm{H} \]\[\phi_A=M\,I_B=(5\times10^{-3}\ \mathrm{H})(1\ \mathrm{A})=5\times10^{-3}\ \mathrm{Wb} \]Answer: \(\displaystyle \phi_A=5\times10^{-3}\ \mathrm{Wb}\)