NCERT Exemplar · Class 12 Physics Moving Charges and Magnetism

(d)Same helix traversed in opposite senses means equal and opposite angular velocities. Opposite senses need opposite-sign charges, so (b) fails; masses stay free, so (a) and (c) are not forced.\[\boldsymbol{\omega}=-\frac{q}{m}\,\mathbf{B}_0 \] \[\boldsymbol{\omega}_1=-\boldsymbol{\omega}_2\;\Rightarrow\;\frac{q_1}{m_1}\,\mathbf{B}_0=-\frac{q_2}{m_2}\,\mathbf{B}_0 \] \[\left(\frac{e}{m}\right)_1+\left(\frac{e}{m}\right)_2=0 \]

(a)The cross product makes \(\displaystyle \mathbf{B}\) perpendicular to \(\displaystyle \mathbf{v}\) and to the line from the electron to the point. Distance enters as \(\displaystyle r^{2}\), not \(\displaystyle r^{3}\).\[\mathbf{B}=\frac{\mu_0}{4\pi}\,\frac{q\,\mathbf{v}\times\hat{\mathbf{r}}}{r^{2}},\qquad q=-e \] \[\mathbf{B}\cdot\mathbf{v}=0,\qquad \mathbf{B}\cdot\hat{\mathbf{r}}=0,\qquad B=\frac{\mu_0}{4\pi}\,\frac{e\,v\sin\theta}{r^{2}} \]

(a)The fixed half gives \(\displaystyle M/2\) along \(\displaystyle \hat{\mathbf{k}}\); the bent half gives \(\displaystyle M/2\) along \(\displaystyle -\hat{\mathbf{i}}\), perpendicular to it. The resultant moment is smaller, and so is the far axial field.\[M=I\pi R^{2},\qquad \mathbf{m}'=\frac{M}{2}\left(\hat{\mathbf{k}}-\hat{\mathbf{i}}\right) \] \[M'=\frac{M}{2}\sqrt{1^{2}+1^{2}}=\frac{M}{\sqrt{2}}<M \] \[\mathbf{B}=\frac{\mu_0}{4\pi z^{3}}\left[3(\mathbf{m}\cdot\hat{\mathbf{k}})\,\hat{\mathbf{k}}-\mathbf{m}\right]\quad (z\gg R) \] \[B=\frac{\mu_0}{4\pi}\,\frac{2M}{z^{3}},\qquad B'=\frac{\mu_0}{4\pi z^{3}}\,\frac{M}{2}\left|2\hat{\mathbf{k}}+\hat{\mathbf{i}}\right|=\frac{\mu_0}{4\pi}\,\frac{\sqrt{5}\,M}{2z^{3}} \] \[\frac{B'}{B}=\frac{\sqrt{5}}{4}\approx 0.56<1 \]

(d)Inside a long solenoid \(\displaystyle \mathbf{B}\) lies along the axis, parallel to the electron's velocity, so the magnetic force is zero: no deflection, no speed change.\[\mathbf{F}=-e\,\mathbf{v}\times\mathbf{B} \] \[F=e\,vB\sin 0^{\circ}=0 \] \[\mathbf{a}=\frac{\mathbf{F}}{m}=0\;\Rightarrow\;\mathbf{v}\ \text{unchanged} \]

(a)Electric field accelerates it between the dees; inside a dee only the magnetic force acts, so speed is constant but direction changes. Acceleration never vanishes, and (b), (c), (d) fail.\[\text{gap:}\quad a=\frac{qE}{m},\qquad \Delta K=qV \] \[\text{dee:}\quad dW=\mathbf{F}_B\cdot\mathbf{v}\,dt=0\;\Rightarrow\;v\ \text{constant} \] \[a=\frac{v^{2}}{r}=\frac{qvB}{m}\neq 0 \]

(d)Turning the loop about the axis perpendicular to its plane turns it about \(\displaystyle \mathbf{M}\) itself. \(\displaystyle \mathbf{M}\) does not move, so its angle with \(\displaystyle \mathbf{B}\) and the energy are unchanged.\[U=-\mathbf{M}\cdot\mathbf{B}=-MB\cos\theta \] \[\theta_f=\theta_i\;\Rightarrow\;W=U_f-U_i=0 \] \[\boldsymbol{\tau}\cdot\hat{\mathbf{M}}=(\mathbf{M}\times\mathbf{B})\cdot\hat{\mathbf{M}}=0 \]