Explanation:
$\overrightarrow{\text{F}}_\text{AD}+\overrightarrow{\text{F}}_\text{BC}=0$
$\overrightarrow{\text{F}}_\text{AB}>\overrightarrow{\text{F}}_\text{CD}$
Force acting on the wire per unit length carrying current i2 due to the wire carrying current i1 placed at a distance d is given by,
$\text{F}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}}$
So, forces per unit length acting on sides AB and CD are as follows:
$\text{F}_\text{AB}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}}$ (Towards the wire)
$\text{F}_\text{CD}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi(\text{d}+\text{a})}$ (Away from the wire)
Here, FAB > FCD because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current i1.
The forces per unit length acting on sides BC and DA will be equal and opposite, as they are equally away from the wire carrying current i1, with current i2 flowing in the opposite direction.
$\therefore\text{F}_\text{BC}=-\text{F}_\text{DA}$
Now,
Net force:
$\text{F} = \text{F}_\text{AB}+\text{F}_\text{BC}+\text{F}_\text{CD}+\text{F}_\text{DA}$
$\Rightarrow\text{F}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}}+\text{F}_\text{BC}-\frac{\mu_0\text{i}_1\text{i}_2}{2\pi(\text{d+a})}-\text{F}_\text{BC}$
$\Rightarrow\text{F}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi}\Big(\frac{1}{\text{d}}-\frac{1}{\text{d+a}}\Big)$
$\Rightarrow\text{F}=\frac{\mu_0\text{i}_1\text{i}_2\text{a}}{2\pi\text{d}(\text{d+a})}$
(Towards the wire)
Therefore, the loop will move towards the wire.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
In a Young’s double slit experiment, the source illuminating the slits is changed from blue to violet. The width of the fringes
|
(a) Increases |
(b) Decreases |
(c) Becomes unequal |
(d) Remains constant |
$\text{F = Mg}$
$\text{F}=\mu\text{Mg}$
$\text{Mg}\leq\text{F}\leq\text{Mg}\sqrt{1+\mu^2}$
$\text{Mg}\geq\text{F}\geq\text{Mg}\sqrt{1-\mu^2}.$
The time period of a freely suspended magnet is 2 sec. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be
| (a) 4 sec | (b) 2 sec |
(c) 
|
(d) 1 sec |
Compton effect is associated with
| (a) α - rays | (b) β - rays | (c) X-rays | (d) Positive rays |
The introduction of a grid in a triode valve affects plate current by
|
(a) Making the thermionic emission easier at low temperature |
|
(b) Releasing more electrons from the plate |
|
(c) By increasing plate voltage |
|
(d) By neutralising space charge |
A positively charged thin metal ring of radius R is fixed in the xy - plane with its centre at the O. A negatively charged particle P is released from rest at the point (0, 0, 
where 
. Then the motion of P is
|
(a) Periodic for all values of |
|
(b) Simple harmonic for all values of satisfying 0 < |
|
(c) Approximately simple harmonic provided |
|
(d) a, c |
