Correct answer option is (A)

$\begin{array}{rl}& \overrightarrow{\mathrm{M}}=\mathrm{i}\overrightarrow{\mathrm{A}}\\ & =5\times (0.2)\times (0.1)(-\hat{\mathrm{i}})\\ & =0.1(-\hat{\mathrm{i}})\\ & \overrightarrow{\tau }=\overrightarrow{\mathrm{M}}\times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}})\times (2\times {10}^{-3})(\hat{\mathrm{j}})\\ & =2\times {10}^{-4}(-\hat{\mathrm{k}})\mathrm{N}-\mathrm{m}\end{array}$

Correct answer option is (B)

Note : Direction of magnetic field is in $+\hat{k}$

$\begin{array}{rl}& \overrightarrow{\mathrm{F}}=\mathrm{i}\overrightarrow{\ell }\times \overrightarrow{\mathrm{B}}\\ & \ell =2\mathrm{R}\\ & \overrightarrow{\mathrm{F}}=-2\mathrm{iRB}\hat{\mathrm{j}}\end{array}$

Correct answer 28

Magnetic moment is defined as the product of the magnet's pole strength and the distance between the poles (also known as the magnetic length). When a magnetic strip is bent, its magnetic moment changes based on the new configuration.

Consider a straight magnetic strip with a magnetic moment of $44{\text{Am}}^2$. If this strip is bent into a semicircular shape, we need to find the new effective magnetic moment.

The magnetic moment in a straight strip is given by:

$M_{\text{straight}}=m\cdot l$

where:

• $m$ is the pole strength
• $l$ is the magnetic length

Given $M_{\text{straight}}=44{\text{Am}}^2$, let's now consider the strip bent into a semicircular shape.

When the strip is bent into a semicircle, the effective distance between the magnetic poles is the diameter of the semicircle. Let's denote the original length of the strip as $L$. In a straight line, this length $L$ is also the magnetic length. When bent into a semicircle, the length of the arc of the semicircle is still $L$.

The circumference of a full circle is given by:

$C=2\pi R$

Therefore, the length of the arc of a semicircle is:

$L=\pi R$

Solving for $R$, we get:

$R=\frac{L}{\pi }$

The diameter of the semicircle (which is the new effective magnetic length, $l_{\text{new}}$) is twice the radius:

$l_{\text{new}}=2R=2\cdot \frac{L}{\pi }=\frac{2L}{\pi }$

Now, the new magnetic moment $M_{\text{new}}$ is:

$M_{\text{new}}=m\cdot l_{\text{new}}=m\cdot \frac{2L}{\pi }$

We know from the original magnetic strip:

$M_{\text{straight}}=m\cdot L=44{\text{Am}}^2$

Rewriting $m$ in terms of the known magnetic moment of the straight strip:

$m=\frac{44}{L}$

Substituting $m$ into the new magnetic moment equation:

$M_{\text{new}}=\frac{44}{L}\cdot \frac{2L}{\pi }$

Canceling out $L$ from the numerator and the denominator:

$M_{\text{new}}=\frac{44\cdot 2}{\pi }=\frac{88}{\pi }$

Given $\pi =\frac{22}{7}$, we substitute this value into the equation:

$M_{\text{new}}=\frac{88\cdot 7}{22}=28{\text{Am}}^2$

Therefore, the magnetic moment of the strip when bent into a semicircular shape is $28{\text{Am}}^2$.

Correct answer is 160

Due to constant component of magnetic field $F=0$

Due to variable component

$\begin{array}{rl}& F_1=0\\ & \text{ and, }{F}_2+F_3=0\\ & \begin{array}{rl}\text{ and, }{F}_4& =\left({\mathrm{B}}_0{4}_x\right)i\mathrm{L}\\ & =5\times 4\times 2\times 2\times 2\\ & =160\mathrm{N}\end{array}\end{array}$

Correct answer is 3

$\begin{array}{rl}& A=36\times {10}^{-4}{\mathrm{m}}^2\\ & I=6\times {10}^{-2}\mathrm{m}\\ & =6\mathrm{cm}\\ V& =\frac{6\mathrm{cm}}{1\sec }=6\mathrm{cm}/\mathrm{s}\\ \epsilon & =B/{\mathrm{vn}}^2=0.5\times \frac{6}{100}\times \frac{6}{100}\\ & =18\times {10}^{-4}\mathrm{V}\\ E& =\frac{n^2{\epsilon }^2}{R}t=100\times \frac{18\times 18\times {10}^{-4}\times {10}^{-4}}{{10}^2}\times 1\\ & =324\times {10}^{-10}\times {10}^2\\ & =3.24\times {10}^{-6}\end{array}$