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NCERT Exemplar · Class 12 Physics Moving Charges and Magnetism

29 questions

Short Answer Questions 17–22 (part 4 of 5)

  1. Exercise 17

    A current carrying loop consists of 3\displaystyle 3 identical quarter circles of radius R\displaystyle R, lying in the positive quadrants of the x−y,y−z\displaystyle x-y, y-z and z−x\displaystyle z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B\displaystyle \mathbf{B} at the origin.
    NCERT’s answer
    \(\displaystyle \mathbf{B}=\frac{1}{4}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \frac{\mu_0 I}{2 R}\)
    Current circulates \(\displaystyle x\to y\to z\to x\) along the three arcs.NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q17Each arc is a quarter of a circular loop, so at the common centre \[B_1=\tfrac14\cdot\frac{\mu_0 I}{2R}=\frac{\mu_0 I}{8R} \] Right-hand rule: the \(\displaystyle x\)-\(\displaystyle y\) arc gives \(\displaystyle +\hat{\mathbf k}\), the \(\displaystyle y\)-\(\displaystyle z\) arc gives \(\displaystyle +\hat{\mathbf i}\), the \(\displaystyle z\)-\(\displaystyle x\) arc gives \(\displaystyle +\hat{\mathbf j}\). \[\mathbf B=\frac{\mu_0 I}{8R}\left(\hat{\mathbf i}+\hat{\mathbf j}+\hat{\mathbf k}\right) \] \[|\mathbf B|=\frac{\mu_0 I}{8R}\sqrt{1^2+1^2+1^2}=\frac{\sqrt3\,\mu_0 I}{8R} \] \[\cos\alpha=\frac{1}{\sqrt3}\;\Rightarrow\;\alpha\approx54.7^\circ\ \text{with each axis} \] Caution: the three fields are mutually perpendicular, so add them as vectors (\(\displaystyle \sqrt3\), not $\displaystyle 3$). Reversing the current reverses \(\displaystyle \mathbf B\).Answer: \(\displaystyle \mathbf B=\dfrac{\mu_0 I}{8R}(\hat{\mathbf i}+\hat{\mathbf j}+\hat{\mathbf k})\), magnitude \(\displaystyle \dfrac{\sqrt3\,\mu_0 I}{8R}\), along the line from the origin towards \(\displaystyle (1,1,1)\).

  2. Exercise 18

    A charged particle of charge e\displaystyle e and mass m\displaystyle m is moving in an electric field E\displaystyle \mathbf{E} and magnetic field B\displaystyle \mathbf{B}. Construct dimensionless quantities and quantities of dimension [T]−1\displaystyle [T]^{-1}.
    NCERT’s answer
    No dimensionless quantity \(\displaystyle [T]^{-1}=[\omega]=\left[\frac{e B}{m}\right]\)
    Charge is an independent dimension \(\displaystyle \mathrm{Q}\): \[[e]=\mathrm{Q},\quad [m]=\mathrm{M},\quad [E]=\frac{[F]}{[q]}=\mathrm{M\,L\,T^{-2}\,Q^{-1}},\quad [B]=\frac{[F]}{[q][v]}=\mathrm{M\,T^{-1}\,Q^{-1}} \] Take \(\displaystyle \Pi=e^{a}m^{b}E^{c}B^{d}\): \[[\Pi]=\mathrm{Q}^{\,a-c-d}\;\mathrm{M}^{\,b+c+d}\;\mathrm{L}^{\,c}\;\mathrm{T}^{\,-2c-d} \] Dimensionless: every exponent zero. \[\mathrm{L}:\ c=0,\qquad \mathrm{T}:\ -d=0\Rightarrow d=0,\qquad \mathrm{Q}:\ a=0,\qquad \mathrm{M}:\ b=0 \] Only the trivial solution: none can be built from \(\displaystyle e,m,E,B\).Dimension \(\displaystyle \mathrm{T^{-1}}\): \[\mathrm{L}:\ c=0,\qquad \mathrm{T}:\ -d=-1\Rightarrow d=1,\qquad \mathrm{Q}:\ a-1=0\Rightarrow a=1,\qquad \mathrm{M}:\ b+1=0\Rightarrow b=-1 \] \[\left[\frac{eB}{m}\right]=\frac{\mathrm{Q}\cdot\mathrm{M\,T^{-1}\,Q^{-1}}}{\mathrm{M}}=\mathrm{T^{-1}} \] Caution: charge must be counted as \(\displaystyle \mathrm{Q}\), or \(\displaystyle a\) cannot be fixed. With the particle's speed \(\displaystyle v\) added, \(\displaystyle \left[\dfrac{vB}{E}\right]=\dfrac{\mathrm{L\,T^{-1}}}{\mathrm{L\,T^{-1}}}=1\).Answer: no dimensionless quantity from \(\displaystyle e,m,E,B\); dimension \(\displaystyle \mathrm{T^{-1}}\): \(\displaystyle \dfrac{eB}{m}\).

  3. Exercise 19

    An electron enters with a velocity v=v0i\displaystyle \mathbf{v}=v_0 \mathbf{i} into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the x−y\displaystyle x-y plane. Suggest a configuration of fields E\displaystyle \mathbf{E} and B\displaystyle \mathbf{B} that can lead to it.
    NCERT’s answer
    \(\displaystyle \mathbf{E}=E_0 \hat{\mathbf{i}}, E_0>0, \mathbf{B}=B_0 \hat{\mathbf{k}}\)
    Take \(\displaystyle \mathbf B\) perpendicular to the plane of the orbit and \(\displaystyle \mathbf E\) in that plane, along the entry direction: \[\mathbf B=B_0\hat{\mathbf k},\qquad \mathbf E=E_0\hat{\mathbf i}\qquad(B_0,E_0>0) \]NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q19Magnetic force on the electron at entry: \[\mathbf F_B=-e\,(v_0\hat{\mathbf i})\times(B_0\hat{\mathbf k})=+e\,v_0B_0\,\hat{\mathbf j} \] In the \(\displaystyle x\)-\(\displaystyle y\) plane and perpendicular to \(\displaystyle \mathbf v\): circular motion, \[r=\frac{mv}{eB_0} \] Electric force: \[\mathbf F_E=-eE_0\,\hat{\mathbf i} \] It opposes the entry motion, doing negative work: \(\displaystyle v\downarrow\Rightarrow r\downarrow\), the circle tightens.Neither force has a \(\displaystyle z\)-component, so \(\displaystyle v_z=0\) throughout.Caution: \(\displaystyle E_z=0\) and \(\displaystyle B_x=B_y=0\) are essential, else the path becomes a helix.Answer: \(\displaystyle \mathbf B\) along \(\displaystyle z\), \(\displaystyle \mathbf E\) along \(\displaystyle +x\).

  4. Exercise 20

    Do magnetic forces obey Newton's third law. Verify for two current elements dl1=dli^\displaystyle \boldsymbol{d} \boldsymbol{l}_{\mathbf{1}}=d l \hat{\mathbf{i}} located at the origin and dl2=dlj^\displaystyle \boldsymbol{d} \boldsymbol{l}_{\mathbf{2}}=d l \hat{\mathbf{j}} located at (0,R,0)\displaystyle (0, R, 0). Both carry current I\displaystyle I.
    NCERT’s answer
    Force due to \(\displaystyle d \mathbf{l}_{\mathbf{2}}\) on \(\displaystyle d \mathbf{l}_{\mathbf{1}}\) is zero. Force due to \(\displaystyle d \mathbf{l}_{\mathbf{1}}\) on \(\displaystyle d \mathbf{l}_{\mathbf{2}}\) is non-zero.
    Field of a current element (Biot–Savart): \[d\mathbf B=\frac{\mu_0}{4\pi}\,\frac{I\,d\mathbf l\times\hat{\mathbf r}}{r^2} \]NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q20Element $\displaystyle 1$ acting on element $\displaystyle 2$ (\(\displaystyle \hat{\mathbf r}=\hat{\mathbf j}\), \(\displaystyle r=R\)): \[d\mathbf B_1=\frac{\mu_0 I\,dl}{4\pi R^2}\,(\hat{\mathbf i}\times\hat{\mathbf j})=\frac{\mu_0 I\,dl}{4\pi R^2}\,\hat{\mathbf k} \] \[d\mathbf F_2=I\,dl\,\hat{\mathbf j}\times d\mathbf B_1=\frac{\mu_0 I^2\,dl^2}{4\pi R^2}\,(\hat{\mathbf j}\times\hat{\mathbf k})=\frac{\mu_0 I^2\,dl^2}{4\pi R^2}\,\hat{\mathbf i} \] Element $\displaystyle 2$ acting on element $\displaystyle 1$ (\(\displaystyle \hat{\mathbf r}=-\hat{\mathbf j}\)): \[d\mathbf B_2\propto\hat{\mathbf j}\times(-\hat{\mathbf j})=\mathbf 0\;\Rightarrow\;d\mathbf F_1=\mathbf 0 \] \[d\mathbf F_1\neq-d\mathbf F_2 \] The third law holds only for complete closed circuits, not for isolated elements.Answer: No: \(\displaystyle d\mathbf F_1=\mathbf 0\) but \(\displaystyle d\mathbf F_2=\dfrac{\mu_0 I^2\,dl^2}{4\pi R^2}\hat{\mathbf i}\).

  5. Exercise 21

    A multirange voltmeter can be constructed by using a galvanometer circuit as shown in Fig. 4.2. We want to construct a voltmeter that can measure 2V, 20V and 200V using a galvanometer of resistance 10Ω\displaystyle 10 \Omega and that produces maximum deflection for current of 1\displaystyle 1 mA. Find R1,R2\displaystyle R_1, R_2 and R3\displaystyle R_3 that have to be used. NCERT_Question_Class12_Physics_Exemplar_Ch4_Q21
    NCERT’s answer
    \(\displaystyle i_{\mathrm{G}}\left(G+R_1\right)=2\) for 2V range \(\displaystyle i_{\mathrm{G}}\left(G+R_1+R_2\right)=20\) for 20V range and \(\displaystyle i_{\mathrm{G}}\left(G+R_1+R_2+R_3\right)=200\) for 200V range Gives \(\displaystyle R_1=1990 \Omega\) \(\displaystyle R_2=18 \mathrm{k} \Omega\) and \(\displaystyle R_3=180 \mathrm{k} \Omega\) NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q21_ncert
    At each terminal the galvanometer and the resistors up to that terminal are in series, and full scale is \(\displaystyle I_g=1\ \mathrm{mA}\).\[2\ \mathrm{V}:\quad G+R_1=\frac{2\ \mathrm{V}}{10^{-3}\ \mathrm{A}}=2000\ \Omega\;\Rightarrow\;R_1=2000-10=1990\ \Omega \] \[20\ \mathrm{V}:\quad G+R_1+R_2=\frac{20\ \mathrm{V}}{10^{-3}\ \mathrm{A}}=20000\ \Omega\;\Rightarrow\;R_2=20000-2000=18000\ \Omega \] \[200\ \mathrm{V}:\quad G+R_1+R_2+R_3=\frac{200\ \mathrm{V}}{10^{-3}\ \mathrm{A}}=200000\ \Omega\;\Rightarrow\;R_3=200000-20000=180000\ \Omega \] Caution: \(\displaystyle R_2\) and \(\displaystyle R_3\) are only the extra resistances; subtract the total already in the loop, not just \(\displaystyle G\).Answer: \(\displaystyle R_1=1990\ \Omega\), \(\displaystyle R_2=18\ \mathrm{k\Omega}\), \(\displaystyle R_3=180\ \mathrm{k\Omega}\).

  6. Exercise 22

    A long straight wire carrying current of 25A rests on a table as shown in Fig. 4.3. Another wire PQ of length 1m, mass 2.5\displaystyle 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise? NCERT_Question_Class12_Physics_Exemplar_Ch4_Q22
    NCERT’s answer
    \(\displaystyle F=B I l \operatorname{Sin} \theta=B I l\) \[B=\frac{\mu_{\mathrm{o}} I}{2 \pi h} \] \[F=m g=\frac{\mu_{\mathrm{o}} \mathrm{I}^2 l}{2 \pi h} \] \[\begin{aligned} h= & \frac{\mu_{\mathrm{o}} I^2 l}{2 \pi m g}=\frac{4 \pi \times 10^{-7} \times 250 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8} \\ & =51 \times 10^{-4} \\ & h=0.51 \mathrm{~cm} \end{aligned} \] NCERT_Solution_Class12_Physics_Exemplar_Ch4_Q22_ncert
    Opposite currents repel, so PQ floats where the repulsion equals its weight.\[F=\frac{\mu_0 I_1 I_2}{2\pi h}\,L=\frac{\mu_0 I^2 L}{2\pi h} \] \[mg=\left(2.5\times10^{-3}\ \mathrm{kg}\right)\left(9.8\ \mathrm{m\,s^{-2}}\right)=2.45\times10^{-2}\ \mathrm{N} \] \[F=mg\;\Rightarrow\;h=\frac{\mu_0 I^2 L}{2\pi\,mg}=\frac{\left(2\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\right)(25\ \mathrm{A})^2(1\ \mathrm{m})}{2.45\times10^{-2}\ \mathrm{N}} \] \[h=\frac{1.25\times10^{-4}\ \mathrm{N\,m}}{2.45\times10^{-2}\ \mathrm{N}}\approx5.1\times10^{-3}\ \mathrm{m} \] Caution: convert $\displaystyle 2.5$ g to kg before using \(\displaystyle mg\).Answer: \(\displaystyle h\approx5\ \mathrm{mm}\) above the lower wire.