Biot-Savart's Law and its application Questions in English

(B) For a long cylindrical wire of radius '$a$' carrying current '$I$':
$1$. Inside the wire $(r < a)$,the magnetic field is given by $B = \frac{\mu_0 I r}{2\pi a^2}$. This shows that $B$ is directly proportional to $r$ $(B \propto r)$,resulting in a linear increase from the center to the surface.
$2$. Outside the wire $(r > a)$,the magnetic field is given by $B = \frac{\mu_0 I}{2\pi r}$. This shows that $B$ is inversely proportional to $r$ $(B \propto 1/r)$,resulting in a hyperbolic decrease as distance increases.
$3$. Combining these,the graph shows a linear increase up to $r = a$ and a hyperbolic decrease for $r > a$. This corresponds to Graph $B$.

(A) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula $B = \frac{\mu_0 I}{2\pi r}$.
Given values are current $I = 90 \text{ A}$ and distance $r = 1.5 \text{ m}$.
The permeability of free space is $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting these values into the formula:
$B = \frac{4\pi \times 10^{-7} \times 90}{2\pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5}$
$B = 2 \times 10^{-7} \times 60 = 120 \times 10^{-7} = 1.2 \times 10^{-5} \text{ T}$.
According to the right-hand thumb rule,if the current flows from east to west,the magnetic field at a point above the wire points towards the north.

(B) Let the two wires be parallel and separated by a distance $d = 30 \text{ cm} = 0.3 \text{ m}$.
The current in each wire is $I = 8 \text{ A}$.
Since the currents flow in opposite directions,the magnetic fields produced by both wires at the midpoint (distance $r = d/2 = 0.15 \text{ m}$) will point in the same direction according to the Right-Hand Thumb Rule.
The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$.
For the first wire,$B_1 = \frac{\mu_0 I}{2\pi (d/2)} = \frac{2 \times 10^{-7} \times 8}{0.15} = \frac{16 \times 10^{-7}}{0.15} = \frac{1600}{15} \times 10^{-7} \approx 106.67 \mu \text{T}$.
Since both wires contribute equally in the same direction,the total magnetic field $B_{total} = B_1 + B_2 = 2 \times B_1 = 2 \times 106.67 \mu \text{T} = 213.33 \mu \text{T}$.
Given the options provided,$213.33 \mu \text{T}$ is closest to $200 \mu \text{T}$,but based on standard physics problem sets where $d$ is often adjusted to yield integer results (e.g.,if $d = 0.2 \text{ m}$),$213.33 \mu \text{T}$ is the calculated value. Among the choices,$213.33$ is not present,but $213.33$ is mathematically correct.

(C) For wire $I$: The magnetic field at center $P$ is the sum of the fields due to the two straight segments and the semicircular arc. The field due to a semi-infinite wire at distance $r$ is $B_{straight} = \frac{\mu_0 I}{4\pi r}$. Since there are two such segments,their contribution is $2 \times \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{2\pi r}$. The field due to the semicircular arc is $B_{arc} = \frac{\mu_0 I}{4r}$. Thus,$B_1 = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} (\frac{2}{\pi} + 1) = \frac{\mu_0 I}{4r} (\frac{2+\pi}{\pi})$.
For wire $II$: The magnetic field at center $Q$ is the difference between the field due to the semicircular arc and the field due to the straight segments. The straight segments contribute $B_{straight} = \frac{\mu_0 I}{4\pi r}$ each. However,based on the geometry,the net field is $B_2 = \frac{\mu_0 I}{4r} - \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4r} (1 - \frac{1}{\pi}) = \frac{\mu_0 I}{4r} (\frac{\pi-1}{\pi})$.
Taking the ratio $\frac{B_1}{B_2} = \frac{\frac{2+\pi}{\pi}}{\frac{\pi-1}{\pi}} = \frac{2+\pi}{\pi-1}$.

(A) The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.
Substituting the given values: $B = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6} \text{ T}$.
The magnetic energy density $u$ is given by $u = \frac{B^2}{2\mu_0}$.
$u = \frac{(4\pi \times 10^{-6})^2}{2 \times 4\pi \times 10^{-7}} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5} \text{ J/m}^3$.
The volume of the cube $V$ is $(1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9} \text{ m}^3$.
The magnetic energy stored $U$ is $U = u \times V$.
$U = (2\pi \times 10^{-5}) \times 10^{-9} = 2\pi \times 10^{-14} \text{ J}$.
Using $\pi = 3.14$,$U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} \text{ J}$.
Comparing this with $\alpha \times 10^{-14} \text{ J}$,we get $\alpha = 6.28$.

(A) Inside the conductor $(r < a)$,the magnetic field is given by $B = \frac{\mu_0 I r}{2 \pi a^2}$,which shows that $B \propto r$. This represents a linear increase in the magnetic field from the axis to the surface of the conductor.
Outside the conductor $(r > a)$,the magnetic field is given by $B = \frac{\mu_0 I}{2 \pi r}$,which shows that $B \propto \frac{1}{r}$. This represents a reciprocal decrease in the magnetic field as the distance from the conductor increases.
Graph $(A)$ correctly displays a linear increase followed by a reciprocal decrease. Therefore,option $(A)$ is correct.