Question

Figure shows region containing uniform magnetic field 'B' which is increasing at constant rate of 'K'. A rod of mass 'm' and length 'L' is released from rest in this field such that center of rod is at center of region containing field. Two point charges $q_1$ and $q_2$ of mass 'm' each are attached at ends of rod then soon after releasing rod.

• A. If $q_1=q_2=q$ then acceleration of center c of rod is zero
• B. If $q_1=q$ and $q_2=-q$ acceleration of center c of rod is $\frac{qLK}{6m}$
• C. If $q_1=q_2=q$ acceleration of $q_1$ is $\frac{3qLK}{14m}$
• D. If $q_1=q$ and $q_2=-q$ then acceleration of c is zero

Answer 1

Answer: (A), (B), (C)

The correct options are
A If $q_1=q_2=q$ then acceleration of center c of rod is zero
B If $q_1=q$ and $q_2=-q$ acceleration of center c of rod is $\frac{qLK}{6m}$
C If $q_1=q_2=q$ acceleration of $q_1$ is $\frac{3qLK}{14m}$


Induced electric field E is
$E\times 2\pi \frac{L}{2}=\frac{d}{dt}B\frac{\pi L^2}{4}\to E=\frac{LK}{4}$

If $q_1=q$ & $q_2=-q$, $F_{net}=2qE=\frac{qLK}{2}$
$a=\frac{F_{net}}{3m}=\frac{qLk}{2\times 3m}=\frac{qLk}{6m}$
If $q_1=q_2=q$
At C,
$F_{net}=0$ so $a=0$

Torque at C, ${\tau }_c=2F\times \frac{L}{2}=qEL\text{and}{\tau }_c=I_c\alpha I_c=(\frac{m{L}^2}{12}+\frac{m{L}^2}{4}\times 2)=\frac{7}{12}m{L}^{2}qE\text{/}L=\frac{7}{12}m{L}^{\text{/}2}\alpha \Rightarrow \alpha =\frac{12qE}{7mL}{a}_{q_1}=\frac{L}{2}\alpha =\frac{\text{/}L}{2}.\frac{12qE}{7m\text{/}L}=\frac{6}{7}\frac{q}{m}.\frac{LK}{4}=\frac{3}{14}\frac{qLK}{m}$