Correct Answer is Option (D)

$\begin{array}{rl}& \mathrm{m}\mathrm{g}=\mathrm{i}\ell \mathrm{B}\\ \\ & 250\times {10}^{-3}\times 9.8=\mathrm{i}\times 2\times 0.7\\ \\ & \mathrm{i}=\frac{0.250\times 9.8}{2\times 0.7}=0.250\times 7\\ \\ & \mathrm{i}=1.75\mathrm{A}\end{array}$

Correct Answer is Option (B)

$\overrightarrow{F}=I(\overrightarrow{\ell }\times \overrightarrow{B})$

$\begin{array}{rl}& =I[(L\hat{i})\times (2\hat{i}+3\hat{j}-4\hat{k})]\\ \\ & =I(4L\hat{j}+3L\hat{k})\end{array}$

$|\overrightarrow{F}|=5\text{ IL }$

Correct Answer is Option (D)

Force per unit length between two long current carrying wires $=\frac{{\mu }_0{l}_1{l}_2}{2\pi d}$

$F=\frac{4\pi \times {10}^{-7}\times 5\times 10}{2\pi \times 10\times {10}^{-2}}={10}^{-4}$ Nm^$-$1

$\Rightarrow$ The force is attractive as direction of current in two conductors is same

Hence, required force is $1\times {10}^{-4}$ Nm^$-$1 and it is attractive.

Correct Answer is Option (D)

Given Mass per unit length of a metallic rod is
$\frac{m}{l}$ = 0.5 kg m^-1

From figure, for equilibrium

mg sin 30^o = I$l$B cos 30^o

$\Rightarrow$ I = $\frac{mg}{lB}\tan {30}^{\circ }$

= $\frac{0.5\times 9.8}{0.25\times \sqrt{3}}$ = 11.32 A

Correct Answer is Option (C)

Since same current flowing through both the wires

i_i = i₂ = i

So F₁ = $\frac{{\mu }_0{i}^2}{2\pi d}$ = F₂

$\therefore$ Magnitude of force per unit length on the middle wire 'B'

F_net = $\sqrt{F_1^2+F_2^2}$ = $\frac{{\mu }_0{i}^2}{\sqrt{2}\pi d}$

Correct Answer is Option (C)

Force on arm AB due to current in conductor XY is

$F_1=\frac{{\mu }_0}{4\pi }\frac{2IiL}{\left(L/2\right)}=\frac{{\mu }_{0}Ii}{\pi }$

acting towards XY in the plane of loop.

Force on arm CD due to current in conductor XY is

$F_2=\frac{{\mu }_0}{4\pi }\frac{2IiL}{3\left(L/2\right)}=\frac{{\mu }_{0}Ii}{3\pi }$

acting away from XY in the plane of loop.

$\therefore$ Net force on the loop = F₁ – F₂

$=\frac{{\mu }_{0}Ii}{3\pi }\left[1-\frac{1}{3}\right]=\frac{2}{3}\frac{{\mu }_{0}Ii}{\pi }$

Correct Answer is Option (A)

Here, ${\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{BCDA}=\overrightarrow{0}$

$\Rightarrow {\overrightarrow{F}}_{BCDA}=-{\overrightarrow{F}}_{AB}=-\overrightarrow{F}$

   ($\because$ F_AB = $\overrightarrow{F}$)

Correct Answer is Option (A)

$F_1>F_2$ as $F\propto \frac{1}{d}$, and $F_3$ and $F_4$ are equal and opposite. Hence, the net attraction force will be towards the conductor.

Correct Answer is Option (B)

Here, ${\overrightarrow{F}}_{BC}=\overrightarrow{F}$

$\because {\overrightarrow{F}}_{AB}=0$

The net magnetic force on a current carrying closed loop in a uniform magnetic field is zero.

$\therefore {\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{AC}=0$

$\Rightarrow {\overrightarrow{F}}_{AC}=-{\overrightarrow{F}}_{BC}$    ($\because$ ${\overrightarrow{F}}_{AB}=0$)

Correct Answer is Option (B)

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by $\tau =NBiA\sin \theta$. Torque is maximum when $\theta$ = 90^o,i.e., the plane of the coil is parallel to the field ${\tau }_{max}=NBiA$

Forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ acting on the coil are equal in magnitude and opposite in direction. As the forces ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_2$ have the same line of action their resultant effect on the coil is zero. The two forces ${\overrightarrow{F}}_3$ and ${\overrightarrow{F}}_4$ are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is $\overrightarrow{F}$ , the net force on the remaining three arms of the loop is - F.

Correct Answer is Option (D)

According to the figure the magnitude of force on the segment QM is F₃ – F₁ and PM is F₂.


Therefore, the magnitude of the force on

segment PQ is $\sqrt{{\left(F_3-F_1\right)}^2+F_2^2}$