Correct Answer is Option (C)

The Lorentz force equation is given as:

$\overrightarrow{F}=-q(\overrightarrow{v}\times \overrightarrow{B})$

The electron is moving along the positive x-axis, so its velocity vector is $\overrightarrow{v}=v_x\hat{i}$. The magnetic field is applied parallel to the negative z-axis, so its magnetic field vector is $\overrightarrow{B}=-B_z\hat{k}$.

Now, we can calculate the cross product of the velocity and magnetic field vectors:

$\overrightarrow{v}\times \overrightarrow{B}=(v_x\hat{i})\times (-B_z\hat{k})$

Using the cross product properties, we get:

$\overrightarrow{v}\times \overrightarrow{B}=-v_x{B}_z(\hat{i}\times \hat{k})$

The cross product of $\hat{i}$ and $\hat{k}$ is $-\hat{j}$, so:

$\overrightarrow{v}\times \overrightarrow{B}=-v_x{B}_z(-\hat{j})=v_x{B}_z\hat{j}$

Since the electron has a negative charge, the magnetic force will be in the opposite direction:

$\overrightarrow{F}=-(-e)(v_x{B}_z\hat{j})=e(v_x{B}_z\hat{j})$

As a result, the electron will experience a magnetic force along the negative y-axis.

Additionally, as mentioned earlier, when a charged particle moves through a magnetic field perpendicular to its velocity, it follows a circular path. In this case, the velocity of the electron is along the positive x-axis, and the magnetic field is along the negative z-axis, which are indeed perpendicular to each other. As a result, the electron will move along a circular path in the magnetic field.

Hence, the correct answer is:

(C) B and E only

Correct Answer is 68

Magnetic fields due to both wires will be perpendicular to each other.

$\begin{array}{rl}& B_1=\frac{{\mu }_0{i}_1}{2\pi d}{B}_2=\frac{{\mu }_0{i}_2}{2\pi d}\\ \\ & B_{\text{net }}=\sqrt{B_1^2+B_2^2}=\frac{{\mu }_0}{2\pi d}\sqrt{i_1^2+i_2^2}\\ \\ & =\frac{4\pi \times {10}^{-7}}{2\pi \times (7/\sqrt{2})\times {10}^{-2}}\times \sqrt{{15}^2+8^2}(\text{As }d=\frac{7}{\sqrt{2}}\mathrm{cm})\\ \\ & =68\times {10}^{-6}\mathrm{T}\end{array}$

Correct Answer is Option (A)

Force between two current carrying wire

$=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}\times L$

Here, $I_1$ & $I_2$ are equal

$F=\frac{{\mu }_0{I}^2}{2\pi d}\times L$

$F\propto I^2$

$\frac{F_I}{F_{2I}}=\frac{I^2}{4I^2}=\frac{1}{4}$

Correct Answer is 9

Force on $5\mathrm{cm}$ side $=I\ell B\sin \theta$

$\begin{array}{rl}& =2\times \frac{5}{100}\times 0.75\times \frac{12}{13}\\ \\ & =\frac{9}{130}\\ \\ & \therefore x=9\end{array}$

Correct Answer is Option (C)

For balancing of force

$\therefore F_{loop}=\mathrm{weight}$

$\left(\frac{V}{R}\right)IB=mg$

$\left(\frac{V}{10}\right)\times \frac{10}{100}\times ({10}^3\times {10}^{-4})=\left(\frac{1}{1000}\right)\times 10$

$V=10$ volts

Correct Answer is Option (D)

$\begin{array}{rl}& \text{ Force per unit length between two parallel straight wires }=\frac{{\mu }_0{\mathrm{i}}_1{\mathrm{i}}_2}{2\pi \mathrm{d}}\\ \\ & \frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{\frac{{\mu }_0(10{)}^2}{2\pi (5\mathrm{cm})}}{\frac{{\mu }_0(20{)}^2}{2\pi \left(\frac{5\mathrm{cm}}{2}\right)}}=\frac{1}{8}\\ \\ & \Rightarrow {\mathrm{F}}_2=8{\mathrm{F}}_1\end{array}$

Correct Answer is Option (A)

Length of wire Y is very high compared to length of wire X. So we can assume length of wire Y is infinite compare to wire X.

Magnetic field due to wire y from 5 cm from wire towards wire X is,

$B=\frac{{\mu }_0{I}_1}{2\pi d}=\frac{{\mu }_0\times 3}{2\pi \times 0.05}$

And direction of B is outward.

Now force on wire X due to wire Y,

$F_{YX}=I_{2}Bl$

$=2\times \frac{{\mu }_0\times 3}{2\pi \times 0.05}\times 0.5$

$=1.2\times {10}^{-5}N$

Current flowing in both the wire in same direction so both wire will attract each other with equal force.

$\therefore$ $F_{YX}=F_{XY}$

$\therefore$ $F_{XY}=1.2\times {10}^{-5}N$, attractive force towards wire X.

Correct Answer is Option (C)

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

$=ilB\sin {60}^{\circ }$

$=10\times \frac{5}{100}\times 0.5\times \frac{\sqrt{3}}{2}$

$=0.2165$ N

Correct Answer is Option (A)

$mg\times \frac{1}{\sqrt{2}}=\frac{ilB}{\sqrt{2}}$

$\Rightarrow i=\frac{mg}{Bl}$

$=\frac{0.45\times 10}{0.15}=30A$

Correct Answer is 5

$\frac{dF}{dl}=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}$

So $\frac{2\times {10}^{-7}\times 5\times 5}{d}=\frac{{10}^{-5}}{10\times {10}^{-2}}$

$d=\frac{2\times {10}^{-7}\times 5\times 5}{{10}^{-4}}$

= 50 mm

= 5 cm

Correct Answer is Option (C)

$\frac{dF}{dl}=2\times {10}^{-6}$ N/m $=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}$

$2\times {10}^{-6}=\frac{2\times {10}^{-7}\times x^2}{0.2}$

$x=\sqrt{2}\simeq 1.4$

Correct Answer is Option (C)

Field at P is $=\left(\frac{{\mu }_0\times i_1}{2\pi r_1}-\frac{{\mu }_0{i}_2}{2\pi r_2}\right)\left(-\hat{k}\right)$

$=-\left(\frac{{\mu }_{0}4}{2\pi \times 0.04}-\frac{{\mu }_0\times 2}{2\pi \times 0.06}\right)\hat{k}=-\frac{{\mu }_0\times 200}{6\pi }\hat{k}$

So, force $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

$=3\pi \left(2\hat{i}+3\hat{j}\right)\times \left(-\left(\frac{{\mu }_0\times 200}{6\pi }\hat{k}\right)\right)$

$=3\pi \left(\frac{200{\mu }_0}{3\pi }\hat{j}-\frac{100{\mu }_0}{\pi }\hat{i}\right)$

$=200{\mu }_0\hat{j}-300{\mu }_0\hat{i}$

$=4\pi \times {10}^{-5}(2\hat{j}-3\hat{i})$

So, $x=3$

Correct Answer is 3

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}=MB\sin {90}^{\circ }$

$=MB=\frac{i\sqrt{3}{l}^2}{4}B$

$=\sqrt{3}\times {10}^{-5}$ N $-$ m

Correct Answer is 1

$\overrightarrow{\tau }=\overrightarrow{M_2}\times \overrightarrow{B_1}$

$\tau =M_2{B}_1\sin {90}^{\circ }$

$=1\times \frac{{\mu }_0}{4\pi }\frac{M_1}{{(1)}^3}1$

= 10^$-$7 N.m

Correct Answer is Option (A)

Force on each wire be along radially outward and equal so, it will take the shape of circle and parallel to the field.

Correct Answer is Option (B)

F = BI2a

= $\frac{{\mu }_{0}I}{2\pi r}I\times 2a$

= $\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}$

$\tau$ = Fcos $\theta$ $\times$ 2$a$

= $\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}$ $\times$ $\frac{b}{\sqrt{b^2+a^2}}\times 2a$

$\Rightarrow$ $\tau$ = $\frac{2{\mu }_0{I}^2{a}^{2}b}{\pi \left(a^2+b^2\right)}$