Correct answer option is (D)

Let's analyze each statement in detail to determine the correct answer:

Statement (I) : "When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account." This statement is true. According to physics, particularly when dealing with electromagnetism and Maxwell's equations, the momentum of the electromagnetic field plays a critical role in conserving momentum in systems where electromagnetic forces are at play. In situations where electric and magnetic fields vary with time, they can carry momentum. Thus, for the conservation laws to hold, including Newton's third law which states that for every action, there is an equal and opposite reaction, the momentum carried by the electromagnetic fields must be included. This is essential in scenarios such as radiation pressure where light (which can be considered an electromagnetic wave) exerts pressure on surfaces, thereby imparting momentum.

Statement (II) : "Ampere's circuital law does not depend on Biot-Savart's law." This statement is false. Historically and mathematically, Ampère's Circuital Law and Biot-Savart Law are closely related in the context of classical electromagnetism. Ampère's Law, particularly in its integral form, relates the integrated magnetic field around a closed loop to the electric current passing through the loop. Biot-Savart Law, on the other hand, is used to calculate the magnetic field generated by a current-carrying element at a point in space. Although Ampère's Law can be derived without directly invoking the Biot-Savart Law, the fundamental understanding and derivations of magnetic fields due to currents, as presented in many textbooks and formulations, show that both laws are manifestations of how moving charges produce magnetic fields. Moreover, the formulation of Ampère's Law was later extended by Maxwell (Maxwell's addition) to include the concept of displacement current, linking it more fundamentally to the changing electric fields and closing a conceptual loop that ties it to the broader electromagnetic theory that includes the effects described by the Biot-Savart Law.

Given the above explanations:

Option D (Statement I is true but Statement II is false) is the correct answer.

Correct answer option is (C)

To find the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ distances from the axis of a long straight wire, we use Ampère's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.

Ampère’s Law is given by:

$\oint \overrightarrow{B}\cdot d\overrightarrow{l}={\mu }_0{I}_{\text{enc}}$

where:

$\overrightarrow{B}$ is the magnetic field,

${\mu }_0$ is the permeability of free space,

$I_{\text{enc}}$ is the enclosed current.

For a point inside the wire (at radius $r=a/2$):

The current enclosed by a radius $r$ is proportional to the area of the cross-section at radius $r$.

The area of the cross-section at radius $r$ is given by:

$\pi {\left(\frac{a}{2}\right)}^2=\frac{\pi a^2}{4}$

The total current $I$ is uniformly distributed, thus the current enclosed $I_{\text{enc}}$ at radius $r=\frac{a}{2}$ is:

$I_{\text{enc}}=I\times \frac{\text{Area enclosed}}{\text{Total area}}=I\times \frac{\frac{\pi a^2}{4}}{\pi a^2}=\frac{I}{4}$

Applying Ampère’s Law inside the conductor, we get:

$B\cdot 2\pi \left(\frac{a}{2}\right)={\mu }_0\left(\frac{I}{4}\right)$

So,

$B\cdot \pi a=\frac{{\mu }_{0}I}{4}$

Therefore, the magnetic field inside the wire at $r=\frac{a}{2}$ is:

$B_{\frac{a}{2}}=\frac{{\mu }_{0}I}{4\pi a}$

For a point outside the wire (at radius $r=2a$):

The total current enclosed by a radius $r=2a$ is the entire current $I$.

Applying Ampère’s Law outside the conductor, we get:

$B\cdot 2\pi (2a)={\mu }_{0}I$

So,

$B\cdot 4\pi a={\mu }_{0}I$

Therefore, the magnetic field outside the wire at $r=2a$ is:

$B_{2a}=\frac{{\mu }_{0}I}{4\pi a}$

Hence, the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ is:

$\frac{B_{\frac{a}{2}}}{B_{2a}}=\frac{\frac{{\mu }_{0}I}{4\pi a}}{\frac{{\mu }_{0}I}{4\pi a}}=1:1$

So, the correct option is:

Option C: $1:1$

Correct answer option is (B)

Let's dig into how the magnetic field behaves in a coaxial cable in relation to the given options. The principle to consider here is Ampère's law, which states that the magnetic field in a loop surrounding a current is proportional to the amount of current enclosed. When we apply this to a coaxial cable, we must look at different regions within the cable.

Option A: Inside the outer conductor

The magnetic field within the outer conductor is not zero because the current in the outer conductor itself contributes to the magnetic field in that region. However, considering the symmetric distribution of current and the geometry of the coaxial cable, there may be a varying magnetic field within the conductor depending on the distance from the center axis.

Option B: Outside the cable

Outside the coaxial cable, the net current enclosed by a path enclosing both conductors is zero because the current in the inner conductor flows in the opposite direction to the equally magnitude current in the outer conductor. These currents being equal and opposite in direction cancel each other out, leading to a net enclosed current of zero. According to Ampère's law, if the net enclosed current is zero, the magnetic field in that space is also zero. Therefore, the magnetic field is zero outside the cable.

Option C: In between the two conductors

In the space between the two conductors, the magnetic field is not zero. This region only encloses the current from the inner conductor. The magnetic field in this region is due to the current in the inner conductor and follows the right-hand rule, which would result in concentric circles of magnetic field around the inner conductor. Since only the inner conductor's current contributes to the magnetic field in this space, Ampère's law suggests that there is a non-zero magnetic field in this region.

Option D: Inside the inner conductor

Within the inner conductor, the magnetic field is not necessarily zero. Like within the outer conductor, the magnetic field inside the inner conductor will depend on the distribution of the current within that conductor. Utilizing the formula derived from Ampère's law for a cylindrical conductor with a uniform current distribution, the magnetic field inside the conductor increases linearly with the distance from the center axis up to the conductor's surface.

In conclusion, the correct option, based on Ampère's law and the principle that the net current enclosed determines the magnetic field outside the current's path, is:

Option B: Outside the cable

This is because the equal and opposite currents in the inner and outer conductors cancel each other, leading to a net magnetic field of zero outside the coaxial cable.

Correct answer option is 500

The magnetic field inside a solenoid can be calculated using the formula:

$B={\mu }_{0}nI$

where:

• $B$ is the magnetic field in teslas (T),
• ${\mu }_0$ is the permeability of free space ($4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}$),
• $n$ is the number of turns per unit length of the solenoid (turns/m),
• $I$ is the current in amperes (A).

Given:

• The magnetic field $B=6.28\times {10}^{-3}\mathrm{T}$,
• The current $I=5\mathrm{A}$,
• The length of the solenoid $L=0.5\mathrm{m}$,

First, let's calculate the number of turns per unit length $n$, which is $n=\frac{m}{L}$ where $m$ is the total number of turns and $L$ is the length of the solenoid.

Rearrange the formula for $B$ to solve for $m$:

$B={\mu }_0\frac{m}{L}I$

Therefore,

$m=\frac{BL}{{\mu }_{0}I}$

Substituting the values we have:

$m=\frac{(6.28\times {10}^{-3}\mathrm{T})(0.5\mathrm{m})}{(4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A})(5\mathrm{A})}$

$m=\frac{6.28\times {10}^{-3}\times 0.5}{4\pi \times {10}^{-7}\times 5}$

$m=\frac{6.28\times 0.5\times {10}^{-3}}{20\pi \times {10}^{-7}}$

$m=\frac{3.14\times {10}^{-3}}{20\pi \times {10}^{-7}}$

$m=\frac{3.14\times {10}^{-3}}{20\times 3.14\times {10}^{-7}}$

$m=\frac{1}{20\times {10}^{-4}}$

$m=\frac{1\times {10}^4}{20}$

$m=500$

Therefore, the value of $m$ is 500 turns.