Correct Option is (A)

Time taken to cross the field region

$=\frac{50}{2}=25\mathrm{s}$

At $10\mathrm{s}$ the loop is inside field and flux is not changing.

$\therefore {\epsilon }_{\text{induced }}=0$

Correct Option is (B)

$\begin{array}{rl}\epsilon & =\mathrm{N}\left(\frac{\mathrm{\Delta }\varphi }{\mathrm{t}}\right)\\ \mathrm{\Delta }\varphi & =(\mathrm{\Delta }\mathrm{B})\mathrm{A}\\ {\mathrm{B}}_{\mathrm{i}}& =5000\mathrm{T},\\ {\mathrm{B}}_{\mathrm{f}}& =3000\mathrm{T}\\ \mathrm{d}& =0.02\mathrm{m}\\ \mathrm{r}& =0.01\mathrm{m}\\ \mathrm{\Delta }\varphi & =(\mathrm{\Delta }\mathrm{B})\mathrm{A}\\ & =(2000)\pi (0.01{)}^2=0.2\pi \\ \epsilon & =\mathrm{N}\left(\frac{\mathrm{\Delta }\varphi }{\mathrm{t}}\right)\Rightarrow 22=\mathrm{N}\left(\frac{0.2\pi }{2}\right)\\ \mathrm{N}& =70\end{array}$

Correct Option is (

According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:

$\epsilon =-\frac{d\mathrm{\Phi }}{dt}$

Where $\epsilon$ is the induced emf, and $\mathrm{\Phi }$ is the magnetic flux.

To find the magnetic flux $\mathrm{\Phi }$ through the rectangular loop, we use the formula:

$\mathrm{\Phi }=B\cdot A\cdot \cos (\theta )$

Where:

• $B$ is the magnetic field strength,
• $A$ is the area of the loop, and
• $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.

The area $A$ of the rectangular loop is:

$A=\text{length}\times \text{width}=2.5\mathrm{m}\times 2\mathrm{m}=5{\mathrm{m}}^2$

Given the angle $\theta ={60}^{\circ }$, we can calculate the initial magnetic flux ${\mathrm{\Phi }}_{initial}$:

${\mathrm{\Phi }}_{initial}=B\cdot A\cdot \cos ({60}^{\circ })$

${\mathrm{\Phi }}_{initial}=4\mathrm{T}\cdot 5{\mathrm{m}}^2\cdot \cos ({60}^{\circ })$

${\mathrm{\Phi }}_{initial}=4\cdot 5\cdot \frac{1}{2}=10\mathrm{Wb}$

(Since $\cos ({60}^{\circ })=\frac{1}{2}$)

When the loop is removed from the magnetic field, the final magnetic flux ${\mathrm{\Phi }}_{final}$ is zero, because the loop is no longer within the magnetic field. Thus, the change in magnetic flux $\mathrm{\Delta }\mathrm{\Phi }$ is:

$\mathrm{\Delta }\mathrm{\Phi }={\mathrm{\Phi }}_{final}-{\mathrm{\Phi }}_{initial}=0-10\mathrm{Wb}=-10\mathrm{Wb}$

The loop is removed from the field in $t=10$ seconds, so the rate of change of magnetic flux is:

$\frac{d\mathrm{\Phi }}{dt}=\frac{\mathrm{\Delta }\mathrm{\Phi }}{\mathrm{\Delta }t}=\frac{-10\mathrm{Wb}}{10\mathrm{s}}=-1\mathrm{Wb}/\mathrm{s}$

Now we can find the average induced emf $\epsilon$:

$\epsilon =-\frac{d\mathrm{\Phi }}{dt}$

$\epsilon =-(-1\mathrm{Wb}/\mathrm{s})$

$\epsilon =+1\mathrm{V}$

Therefore, the average induced emf in the loop during this time is $+1\mathrm{V}$. The correct answer is:

Option C

$+1\mathrm{V}$

)

Correct answer is 5

$\begin{array}{rl}& \varphi =\mathrm{NAB}\cos (\omega \mathrm{t})\\ \\ & \epsilon =-\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\mathrm{NAB}\omega \sin (\omega \mathrm{t})\\ \\ & {\epsilon }_{max}=\mathrm{NAB}\omega \\ \\ & =200\times 0.2\times 0.01\times \pi \\ \\ & =\frac{4\pi }{10}=\frac{2\pi }{5}\mathrm{volt}\end{array}$

Correct answer is 2

$\begin{array}{rl}& \epsilon =-\left(\frac{\mathrm{d}\varphi }{\mathrm{dt}}\right)=10\mathrm{t}-36\\ & \text{ at }\mathrm{t}=2,\epsilon =16\mathrm{V}\\ & \mathrm{i}=\frac{\epsilon }{\mathrm{R}}=\frac{16}{8}=2\mathrm{A}\end{array}$