Ampere's Circuit Law MCQ Quiz in తెలుగు - Objective Question with Answer for Ampere's Circuit Law - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Magnetic Intensity (H): 

• The ability of a magnetic field to magnetize a material medium is called its magnetic intensity H.
• Its magnitude is measured by the number of ampere-turns flowing round unit length of a solenoid, required to produce that magnetic field.
• Let the field due to a solenoid of n turns per meter length be H = NI / L, where N is the total number of turns and L is the length of the solenoid.
• H does not depend upon the nature of the medium. It is a vector and is directed along the axis of the solenoid.
• It can also be given as H = B / μ 

Consider A as the magnetic potential vector in the field, then

Magnetic field density B = ∇ × A

Divergence of curl of any vector field is zero, shown as

∇ ⋅ (∇ × A) = 0 

⇒ ∇ ⋅ B = 0

⇒ ∇ ⋅ μ H = 0 ⇒ ∇ ⋅ H = 0

Ampere's Law: 

The curl of the magnetic field is equal to the Electric Current Density.

∇ × H = J

so, ∇ × H ≠ 0

Explanation:

Ampere's Circuit Law

Definition: Ampere's circuit law, also known as Ampere's law, relates the circulating magnetic field in a closed loop to the electric current passing through the loop. It is one of Maxwell's equations and is fundamental in the study of electromagnetism. Mathematically, it is expressed as:

∮_L H · dl = I_enc

where:

• ∮_L H · dl is the line integral of the magnetic field H around a closed loop L.
• I_enc is the total current enclosed by the loop.

Correct Option Analysis:

The correct option is:

Option 1: ∇ × J = H

This option is not a true derivative of Ampere's circuit law. To understand why, let's delve deeper into the mathematical formulation and implications of Ampere's law:

Ampere's law in differential form is given by:

∇ × H = J

This equation states that the curl of the magnetic field H is equal to the current density J. In other words, the magnetic field around a current-carrying conductor is directly related to the current flowing through it. The operator ∇ × represents the curl, which measures the tendency of the field to circulate around a point.

In option 1, ∇ × J = H, the roles of H and J are incorrectly interchanged. The curl of the current density J does not yield the magnetic field H. Therefore, this expression is not consistent with Ampere's law and is incorrect.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: I_enc = ∫_S (∇ × H) · dS

This option is a valid expression derived from Ampere's law. By applying Stokes' theorem, which relates the line integral of a vector field over a closed loop to the surface integral of the curl of the vector field over a surface bounded by the loop, we get:

∮_L H · dl = ∫_S (∇ × H) · dS = I_enc

This shows that the total current enclosed by the loop is equal to the surface integral of the curl of the magnetic field over the surface bounded by the loop.

Option 3: I_enc = ∫_S J · dS

This option is also a valid expression. It states that the total current enclosed by the loop is equal to the surface integral of the current density J over the surface S. This is consistent with the definition of current density and the total current passing through a surface.

Option 4: I_enc = ∮_L H · dl

This option is the integral form of Ampere's law itself. It states that the line integral of the magnetic field H around a closed loop is equal to the total current I_enc enclosed by the loop. This is the fundamental form of Ampere's law and is correct.

Conclusion:

Understanding Ampere's circuit law and its derivatives is crucial in the study of electromagnetism. The law provides a relationship between the magnetic field and the electric current passing through a closed loop. The correct expression of this relationship is essential for accurately describing the behavior of electromagnetic fields. Option 1, ∇ × J = H, is not a true derivative of Ampere's law because it incorrectly interchanges the roles of the magnetic field H and the current density J.

This phenomenon is known as Ampere's force law or the parallel currents interaction. According to the right-hand rule, the magnetic fields produced by the currents in the two conductors will be in the same direction between the conductors and in the opposite direction outside the conductors.

Therefore, when two parallel conductors carry current in opposite directions, they will exert a repulsive force on each other.

Concept:

Ampere's circuital law states that the line integral of the magnetic field surrounding the closed-loop equals the number of times the algebraic sum of currents passing through the loop.

∫ B. dl = μ0 I

where B is the magnetic field, dl is the infinitesimal distance, I is the current.

Calculations:

Let the current density be σ.

r = a/3 and r = 2a

From Ampere's Law

∫ B. dl = μ₀ I

Bdl cos θ = μ₀σπ(2/3)²

∴ B = μσa/6

Similarly for other r

∴ B = μσa/4

Thus from (1) and (2) -

The ratio of the magnetic fields is 2/3      

The correct answer is option (3).

Statement 1:

Biot-savart’s law states that the magnetic intensity at any point due to a steady current in an infinitely long straight wire is directly proportional to the current and inversely proportional to the distance from point to wire. It is a general modification of ampere’s law.

\(dH = \frac{{\vec I \cdot d\vec l \times \widehat {{i_r}}}}{{4\pi {R^2}}} = \frac{{\vec I \cdot d\vec l \times \vec R}}{{4\pi {{\left| {\vec R} \right|}^3}}}\)

\(H = \oint \frac{{\vec I \cdot d\vec l \times \widehat {{i_r}}}}{{4\pi {R^2}}}\)

Statement 2:

Ampere's Law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.

It is used for determining magnetic field intensity for symmetrical current distributions.

Therefore, Statement 1 is FALSE and Statement 2 is TRUE.

By, Biot Savart’s Law, the magnetic field due to each \({\rm{Idl}}\) element will be equal and directed along the axis. We know that, the magnetic field due to circular loop is

\({\rm{H}} = \frac{{\rm{I}}}{{2{\rm{ a}}}}\)

So, due to semicircular loop, the field will be half,

\(\rm \vec J = \vec\nabla \times\vec H = \left| {\begin{array}{*{20}{c}} \rm {{{\vec a}_x}}&\rm {{{\vec a}_y}}&\rm {{{\vec a}_z}}\\ \rm {\partial /\partial x}&\rm {\partial /\partial y}&\rm {\partial /\partial z}\\ \rm {{y^2}z}&\rm {\left( {z + 1} \right)x}&\rm {{z^2}} \end{array}} \right| \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = {\vec a_x}\left( { - x} \right) + {\vec a_y}\left( {{y^2}} \right) + {\vec a_z}\left( {\left( {z + 1} \right) - 2yz} \right)\)

at point \(\rm (0, 1, 1)\),

\(\rm \vec J = {\hat a_x}\left( 0 \right) + {\hat a_y}{\left( 1 \right)^2} + {\hat a_z}\left( {2 - 2\left( 1 \right)\left( 1 \right)} \right) = {\hat a_y}\)

\(\rm {\left. {\vec J} \right|_{\left( {0,1,1} \right)}} = 1.{\hat a_y}\;A/{m^2}\).

\(\rm \vec J = \vec \nabla \times \vec H = \frac{1}{\rho }\left| {\begin{array}{*{20}{c}}\rm {{{\hat a}_\rho }}&\rm {\rho \begin{array}{*{20}{c}} \rm {{{\hat a}_\phi }}&\rm {{{\hat a}_z}} \end{array}}\\ {\begin{array}{*{20}{c}} \rm {\frac{\partial }{{\partial \rho }}}\\ \rm {{H_\rho }} \end{array}}&\rm {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} \rm {\frac{\partial }{{\partial \phi }}}\\ \rm {\rho {H_\phi }} \end{array}}&{\begin{array}{*{20}{c}} \rm {\frac{\partial }{{\partial z}}}\\ \rm {{H_z}} \end{array}} \end{array}} \end{array}} \right| = \frac{1}{\rho }\left| {\begin{array}{*{20}{c}}\rm {{{\hat a}_\rho }}&\rm {\rho \begin{array}{*{20}{c}}\rm {{{\hat a}_\phi }}&\rm {{{\hat a}_z}} \end{array}}\\ {\begin{array}{*{20}{c}} \rm {\frac{\partial }{{\partial \rho }}}\\ 0 \end{array}}&{\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}}\rm {\frac{\partial }{{\partial \phi }}}\\ {\rho \frac{{15}}{\rho }} \end{array}}&{\begin{array}{*{20}{c}} {\frac{\partial }{{\partial z}}}\\ 0 \end{array}} \end{array}} \end{array}} \right|\)

\(\rm \Rightarrow \frac{1}{\rho }\frac{\partial }{{\partial \rho }}\left( {\rho {H_\phi }} \right){\hat a_z}\)

\(\rm \Rightarrow \frac{1}{\rho }\frac{\partial }{{\partial \rho }}\left( {\rho .\frac{{15}}{\rho }} \right) = \frac{1}{\rho }\frac{\partial }{{\partial \rho }}\left( {15} \right) = 0\).