JEE Main 2022 Physics Question Paper with Answer and Solution

(A) The magnetic field at the center of a circular coil with $N$ turns is given by the formula:
$B = \frac{N \mu_{0} I}{2 R}$
Given values:
$N = 100$
$R = 5\,cm = 0.05\,m = 5 \times 10^{-2}\,m$
$B = 37.68 \times 10^{-4}\,T$
$\mu_{0} = 4 \pi \times 10^{-7}\,T \cdot m/A$
$\pi = 3.14$
Substituting the values into the formula:
$37.68 \times 10^{-4} = \frac{100 \times 4 \times 3.14 \times 10^{-7} \times I}{2 \times 5 \times 10^{-2}}$
Simplify the equation:
$37.68 \times 10^{-4} = \frac{400 \times 3.14 \times 10^{-7} \times I}{10 \times 10^{-2}}$
$37.68 \times 10^{-4} = \frac{1256 \times 10^{-7} \times I}{10^{-1}}$
$37.68 \times 10^{-4} = 1256 \times 10^{-6} \times I$
$37.68 \times 10^{-4} = 1.256 \times 10^{-3} \times I$
$I = \frac{37.68 \times 10^{-4}}{1.256 \times 10^{-3}} = \frac{3.768 \times 10^{-3}}{1.256 \times 10^{-3}} = 3\,A$
Thus,the current through the coil is $3\,A$.

(A) The resultant intensity $I_R$ of two interfering beams is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
At point $A$,the phase difference $\phi = 0$. Therefore,the intensity is maximum:
$I_A = I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{9I} + \sqrt{4I})^2 = (3\sqrt{I} + 2\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$.
At point $B$,the phase difference $\phi = \pi$. Therefore,the intensity is minimum:
$I_B = I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{9I} - \sqrt{4I})^2 = (3\sqrt{I} - 2\sqrt{I})^2 = (\sqrt{I})^2 = I$.
The difference between the resultant intensities at points $A$ and $B$ is:
$I_A - I_B = 25I - I = 24I$.

(B) The length of the wire $L = 314 \, cm = 3.14 \, m$.
When the wire is bent into a circle of radius $R$,the circumference is $2 \pi R = L$.
$2 \times 3.14 \times R = 3.14 \implies R = 0.5 \, m$.
The area of the coil is $A = \pi R^2 = 3.14 \times (0.5)^2 = 3.14 \times 0.25 = 0.785 \, m^2$.
The magnetic moment $M$ is given by $M = I \times A$.
$M = 14 \, A \times 0.785 \, m^2 = 10.99 \, A \cdot m^2$.
Rounding to the nearest integer,we get $M \approx 11 \, A \cdot m^2$.

(B) The incident ray vector is $\overrightarrow{A} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}$.
The normal to the $X-Y$ plane is along the $Z$-axis,i.e.,$\hat{k}$.
The angle of incidence $i$ is the angle between the incident ray and the normal. Since the ray is travelling in $M_{1}$ $(Z \geq 0)$ towards the origin,its direction vector is $-\overrightarrow{A} = -4\sqrt{3}\hat{i} + 3\sqrt{3}\hat{j} + 5\hat{k}$.
The cosine of the angle $i$ with the normal $\hat{k}$ is given by $\cos i = \frac{|A_z|}{|\overrightarrow{A}|}$.
Magnitude $|\overrightarrow{A}| = \sqrt{(4\sqrt{3})^2 + (-3\sqrt{3})^2 + (-5)^2} = \sqrt{48 + 27 + 25} = \sqrt{100} = 10$.
$\cos i = \frac{5}{10} = \frac{1}{2} \Rightarrow i = 60^{\circ}$.
Using Snell's Law: $\mu_1 \sin i = \mu_2 \sin r$.
$\sqrt{2} \sin 60^{\circ} = \sqrt{3} \sin r$.
$\sqrt{2} \times \frac{\sqrt{3}}{2} = \sqrt{3} \sin r$.
$\sin r = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow r = 45^{\circ}$.
The difference between the angle of incidence and the angle of refraction is $i - r = 60^{\circ} - 45^{\circ} = 15^{\circ}$.

(B) The electric field intensity $E$ in the depletion region is given by the ratio of the potential barrier $V$ to the width of the depletion layer $d$.
$E = \frac{V}{d}$
Given:
Potential barrier $V = 0.6\,V$
Width of depletion layer $d = 6 \times 10^{-6}\,m$
Substituting the values:
$E = \frac{0.6\,V}{6 \times 10^{-6}\,m}$
$E = 0.1 \times 10^{6}\,V/m$
$E = 1 \times 10^{5}\,V/m$
Since $1\,V/m = 1\,N/C$,the electric field intensity is $1 \times 10^{5}\,N/C$.
Thus,the missing value is $1$.

(C) The angle of incidence $i$ is the angle between the incident ray and the normal. The problem states the light enters at an angle of $45^{\circ}$ with the interface. Therefore,the angle of incidence is $i = 90^{\circ} - 45^{\circ} = 45^{\circ}$.
The angle of deviation $D$ is given as $15^{\circ}$. The relationship between the angle of incidence $i$,the angle of refraction $r$,and the angle of deviation $D$ is $D = i - r$.
Substituting the given values: $15^{\circ} = 45^{\circ} - r$,which gives $r = 45^{\circ} - 15^{\circ} = 30^{\circ}$.
Using Snell's law,$n_1 \sin i = n_2 \sin r$,where $n_1 = 1$ (for air) and $n_2 = \mu$ (refractive index of the medium):
$1 \times \sin 45^{\circ} = \mu \times \sin 30^{\circ}$
$\frac{1}{\sqrt{2}} = \mu \times \frac{1}{2}$
$\mu = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$.
Thus,the refractive index of the medium is $1.414$.