Biot Savart's Law MCQ Quiz - Objective Question with Answer for Biot Savart's Law - Download Free PDF

Explanation:

Magnetic Field Intensity Produced by a Differential Current Element

Definition: The magnetic field intensity (dH) at a point P due to a small current element Idl is a fundamental concept in electromagnetism. This is described by the Biot-Savart Law, which relates the magnetic field produced by a current-carrying conductor to the distance from the conductor and the angle between the current element and the position vector.

Biot-Savart Law: According to the Biot-Savart Law, the differential magnetic field intensity (dH) at a point P due to a differential current element Idl is given by:

\[\rm dH = \frac{I \, dl \, \sin \alpha}{4 \pi R^2}\]

Here,

• I is the current flowing through the differential element.
• dl is the length of the differential current element.
• R is the distance between the point P and the current element.
• α is the angle between the current element and the line joining the point P to the element.

Explanation of Correct Option:

The correct option is:

Option 2: \(\rm dH = \frac{Idl \sin \alpha}{4 \pi R^2}\)

This option correctly applies the Biot-Savart Law to describe the magnetic field intensity produced by the differential current element at point P. The law states that the magnetic field intensity is directly proportional to the current and the length of the current element, inversely proportional to the square of the distance, and depends on the sine of the angle between the element and the line joining the point P.

Derivation and Analysis:

To derive and understand why Option 2 is correct, let’s consider the Biot-Savart Law in more detail.

The Biot-Savart Law in vector form is given by:

\[\vec{dH} = \frac{I \, d\vec{l} \times \hat{r}}{4 \pi R^2}\]

Where:

• \(d\vec{l}\) is the differential length vector of the current element.
• \(\hat{r}\) is the unit vector in the direction from the current element to the point P.
• R is the distance between the point P and the current element.

The cross product \(d\vec{l} \times \hat{r}\) introduces the sine of the angle α between the current element and the line joining the point P to the element. Therefore, the magnitude of the magnetic field intensity can be written as:

\[\rm dH = \frac{I \, dl \, \sin \alpha}{4 \pi R^2}\]

This matches the expression given in Option 2, confirming its correctness.

Additional Information:

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

Option 1: \(\rm dH = \frac{Idl \sin \alpha}{R^2}\)

This option does not include the factor \(4 \pi\) in the denominator, which is essential according to the Biot-Savart Law. The \(4 \pi\) factor arises from the integral form of the law and the symmetry of the magnetic field in three-dimensional space.

Option 3: \(\rm dH = \frac{Idl \sin \alpha}{2 \pi R^2}\)

This option includes a factor of \(2 \pi\) instead of \(4 \pi\). This discrepancy indicates an incorrect application of the Biot-Savart Law. The correct factor is \(4 \pi\), which accounts for the complete solid angle around the current element.

Option 4: \(\rm dH = \frac{Idl \sin \alpha}{4 \pi^2 R^2}\)

This option includes an additional \(\pi\) in the denominator, which is not justified by the Biot-Savart Law. The presence of \(4 \pi^2\) instead of \(4 \pi\) is incorrect and does not align with the standard formulation of the law.

Conclusion:

Understanding the Biot-Savart Law is crucial for accurately determining the magnetic field intensity produced by a current element. The correct expression for the magnetic field intensity at a point P due to a differential current element is:

\(\rm dH = \frac{Idl \sin \alpha}{4 \pi R^2}\)

This formulation highlights the dependence on the current, the length of the current element, the distance to the point, and the angle between the element and the line joining the point P. This fundamental law is a cornerstone in electromagnetism, providing insights into the behavior of magnetic fields around current-carrying conductors.

Explanation:

The magnetic field dB due to a small current segment dl' can be expressed as:
\(dB = \frac{μ_0 I}{4π} \frac{dl' \cos θ}{r^2}, \)

where μ₀ is the permeability of free space, I is the current, and θ is the angle between the current element and the observation point. When integrating dl' around the loop, the magnetic field contribution sweeps out a cone. Due to symmetry, the horizontal components of the magnetic field cancel out, leaving only the vertical components, which combine to give the net field.

Noting  that the vector dl' and the unit vector \(\hat{r} \)are perpendicular, the factor cosθ projects out the vertical component of the field. Additionally, cosθ and r² are constants for all points on the loop, and the integral of dl' around the loop simply gives the circumference, 2πR , where R is the radius of the loop.

Substituting this into the expression for B(z) , we get:

\(B(z) = \frac{μ_0 I}{4π} \left( \frac{\cos θ}{r^2} \right) 2π R. \)

Simplifying further, we use the fact that \(r = \sqrt{R^2 + z^2}\) and\( \cos θ = \frac{R}{\sqrt{R^2 + z^2}}\) , which leads to the final expression for the magnetic field along the axis of the current loop:

\( B(z) = \frac{μ_0 I R^2}{2 (R^2 + z^2)^{3/2}}.\)

The correct answer is 1) :\( B(z) = \frac{μ_0 I R^2}{2 (R^2 + z^2)^{3/2}}.\)

Concept:

According to Biot-Savart’s law, the magnetic field at a point due to the incremental element of length dl carrying a current is given by

\(dB = \frac{{{\mu _0}Idl}}{{4\pi {r^2}}}\)

Where B is a magnetic field in Tesla

I, is the current in the wire in Ampere

r is the radius of the wire in meter

From the above expression, it is clear that the magnetic field at a point due to the incremental element of length dl carrying a current is directly proportional to the current carried by the element.

Magnetic field intensity for Various Current Distribution

For line current

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

Surface current 

\(\vec H = \int\limits_S {\frac{{\vec K \cdot d\vec S \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec K\): Surface current density

Volume current

\(\vec H = \int\limits_V {\frac{{\vec J \cdot d\vec V \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec J\): volume current density

In all lines, surface and volume.

\(\vec H \propto \frac{1}{R^2}\)

The magnetic intensity is inversely proportional to the Square of the distance between them.

Concept:

• Biot Savart Law:
  • Biot-savart’s law gives the magnetic field produced due to the current carrying segment.
  • This segment is taken as a vector quantity known as the current element.

• The magnitude of the magnetic field dB at a distance r from a current-carrying element dl is found to be proportional to I and the length dl.
• Formula, \(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)
• Here, \(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Calculation:

Given,

The length of segment, dl = 2.0i cm = 0.02i m

Current, I = 4 A

 Position vector, \(\vec r= 4.0 \hat j m\)

\(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Biot- savart law in vector form, 

\(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)

\(B=10^{-7} ×4× \frac{0.02\hat i× 4\hat j}{4^3}\)

B = 5.0 × 10^-10k T

Concept:

According to Biot-Savart’s law, the magnetic field at a point due to the incremental element of length dl carrying a current is given by

\(dB = \frac{{{\mu _0}Idl}}{{4\pi {r^2}}}\)

Where B is a magnetic field in Tesla

I, is the current in the wire in Ampere

r is the radius of the wire in meter

From the above expression, it is clear that the magnetic field at a point due to the incremental element of length dl carrying a current is directly proportional to the current carried by the element.

Magnetic field intensity for Various Current Distribution

For line current

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

Surface current 

\(\vec H = \int\limits_S {\frac{{\vec K \cdot d\vec S \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec K\): Surface current density

Volume current

\(\vec H = \int\limits_V {\frac{{\vec J \cdot d\vec V \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec J\): volume current density

In all lines, surface and volume.

\(\vec H \propto \frac{1}{R^2}\)

The magnetic intensity is inversely proportional to the Square of the distance between them.

Concept:

• Biot Savart Law:
  • Biot-savart’s law gives the magnetic field produced due to the current carrying segment.
  • This segment is taken as a vector quantity known as the current element.

• The magnitude of the magnetic field dB at a distance r from a current-carrying element dl is found to be proportional to I and the length dl.
• Formula, \(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)
• Here, \(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Calculation:

Given,

The length of segment, dl = 2.0i cm = 0.02i m

Current, I = 4 A

 Position vector, \(\vec r= 4.0 \hat j m\)

\(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Biot- savart law in vector form, 

\(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)

\(B=10^{-7} ×4× \frac{0.02\hat i× 4\hat j}{4^3}\)

B = 5.0 × 10^-10k T

Concept:

According to Biot-Savart’s law, the magnetic field at a point due to the incremental element of length dl carrying a current is given by

\(dB = \frac{{{\mu _0}Idl}}{{4\pi {r^2}}}\)

Where B is a magnetic field in Tesla

I, is the current in the wire in Ampere

r is the radius of the wire in meter

From the above expression, it is clear that the magnetic field at a point due to the incremental element of length dl carrying a current is directly proportional to the current carried by the element.

Magnetic field intensity for Various Current Distribution

For line current

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

Surface current 

\(\vec H = \int\limits_S {\frac{{\vec K \cdot d\vec S \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec K\): Surface current density

Volume current

\(\vec H = \int\limits_V {\frac{{\vec J \cdot d\vec V \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec J\): volume current density

In all lines, surface and volume.

\(\vec H \propto \frac{1}{R^2}\)

The magnetic intensity is inversely proportional to the Square of the distance between them.

According to Biot-Savart law,

\(B=\frac{{{\mu }_{0}}}{4\pi }I\frac{dl\times \hat{r}}{{{r}^{2}}}\)

The direction of magnetic field B is the direction of current element dl crossed into the radial direction of the wire , which is in the opposite direction.

The coils are identical and I₁ = I₂

⇒ B₁ = B₂ and hence,

 \(\vec{B}={{\vec{B}}_{1}}+\left( -{{{\vec{B}}}_{2}} \right)=0\)

Concept:

The magnetic field at a distance ‘R’ from a current-carrying conductor carrying current I is given by:

\(B = \frac{{{\mu _0}I}}{{2\pi R}}\)

Application:

Since the two wires are carrying current in the opposite direction, the magnetic field due to wire ‘1’ will be:

\({B_1} = \frac{{{\mu _0}I}}{{2\pi r}}\)

The direction, according the right-hand rule, will be downwards into the plane of the paper/screen.

Magnetic field due to wire ‘2’

Similarly \({B_2} = \frac{{{\mu _0}\left( {2I} \right)}}{{2\pi \left( {3r} \right)}}\)

\(B_2= \frac{{2{\mu _0}I}}{{2\pi 3r}}\) 

The direction for this is also into the plane of the paper.

Adding the two we get:

B = B₁ + B₂

\(= \frac{{{U_0}I}}{{2\pi r}} + \frac{{2{\mu _0}I}}{{2\pi 3r}}\)

\(= \frac{{{u_0}I}}{{2\pi r}}\left[ {1 + \frac{2}{3}} \right]\)

Explanation:

Magnetic Field Intensity Produced by a Differential Current Element

Definition: The magnetic field intensity (dH) at a point P due to a small current element Idl is a fundamental concept in electromagnetism. This is described by the Biot-Savart Law, which relates the magnetic field produced by a current-carrying conductor to the distance from the conductor and the angle between the current element and the position vector.

Biot-Savart Law: According to the Biot-Savart Law, the differential magnetic field intensity (dH) at a point P due to a differential current element Idl is given by:

\[\rm dH = \frac{I \, dl \, \sin \alpha}{4 \pi R^2}\]

Here,

• I is the current flowing through the differential element.
• dl is the length of the differential current element.
• R is the distance between the point P and the current element.
• α is the angle between the current element and the line joining the point P to the element.

Explanation of Correct Option:

The correct option is:

Option 2: \(\rm dH = \frac{Idl \sin \alpha}{4 \pi R^2}\)

This option correctly applies the Biot-Savart Law to describe the magnetic field intensity produced by the differential current element at point P. The law states that the magnetic field intensity is directly proportional to the current and the length of the current element, inversely proportional to the square of the distance, and depends on the sine of the angle between the element and the line joining the point P.

Derivation and Analysis:

To derive and understand why Option 2 is correct, let’s consider the Biot-Savart Law in more detail.

The Biot-Savart Law in vector form is given by:

\[\vec{dH} = \frac{I \, d\vec{l} \times \hat{r}}{4 \pi R^2}\]

Where:

• \(d\vec{l}\) is the differential length vector of the current element.
• \(\hat{r}\) is the unit vector in the direction from the current element to the point P.
• R is the distance between the point P and the current element.

The cross product \(d\vec{l} \times \hat{r}\) introduces the sine of the angle α between the current element and the line joining the point P to the element. Therefore, the magnitude of the magnetic field intensity can be written as:

\[\rm dH = \frac{I \, dl \, \sin \alpha}{4 \pi R^2}\]

This matches the expression given in Option 2, confirming its correctness.

Additional Information:

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

Option 1: \(\rm dH = \frac{Idl \sin \alpha}{R^2}\)

This option does not include the factor \(4 \pi\) in the denominator, which is essential according to the Biot-Savart Law. The \(4 \pi\) factor arises from the integral form of the law and the symmetry of the magnetic field in three-dimensional space.

Option 3: \(\rm dH = \frac{Idl \sin \alpha}{2 \pi R^2}\)

This option includes a factor of \(2 \pi\) instead of \(4 \pi\). This discrepancy indicates an incorrect application of the Biot-Savart Law. The correct factor is \(4 \pi\), which accounts for the complete solid angle around the current element.

Option 4: \(\rm dH = \frac{Idl \sin \alpha}{4 \pi^2 R^2}\)

This option includes an additional \(\pi\) in the denominator, which is not justified by the Biot-Savart Law. The presence of \(4 \pi^2\) instead of \(4 \pi\) is incorrect and does not align with the standard formulation of the law.

Conclusion:

Understanding the Biot-Savart Law is crucial for accurately determining the magnetic field intensity produced by a current element. The correct expression for the magnetic field intensity at a point P due to a differential current element is:

\(\rm dH = \frac{Idl \sin \alpha}{4 \pi R^2}\)

This formulation highlights the dependence on the current, the length of the current element, the distance to the point, and the angle between the element and the line joining the point P. This fundamental law is a cornerstone in electromagnetism, providing insights into the behavior of magnetic fields around current-carrying conductors.

Concept:

• Biot Savart Law:
  • Biot-savart’s law gives the magnetic field produced due to the current carrying segment.
  • This segment is taken as a vector quantity known as the current element.

• The magnitude of the magnetic field dB at a distance r from a current-carrying element dl is found to be proportional to I and the length dl.
• Formula, \(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)
• Here, \(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Calculation:

Given,

The length of segment, dl = 2.0i cm = 0.02i m

Current, I = 4 A

 Position vector, \(\vec r= 4.0 \hat j m\)

\(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Biot- savart law in vector form, 

\(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)

\(B=10^{-7} ×4× \frac{0.02\hat i× 4\hat j}{4^3}\)

B = 5.0 × 10^-10k T

Concept:

According to Biot-Savart’s law, the magnetic field at a point due to the incremental element of length dl carrying a current is given by

\(dB = \frac{{{\mu _0}Idl}}{{4\pi {r^2}}}\)

Where B is a magnetic field in Tesla

I, is the current in the wire in Ampere

r is the radius of the wire in meter

From the above expression, it is clear that the magnetic field at a point due to the incremental element of length dl carrying a current is directly proportional to the current carried by the element.

Magnetic field intensity for Various Current Distribution

For line current

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

Surface current 

\(\vec H = \int\limits_S {\frac{{\vec K \cdot d\vec S \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec K\): Surface current density

Volume current

\(\vec H = \int\limits_V {\frac{{\vec J \cdot d\vec V \times \widehat {{i_r}}}}{{4\pi {R^2}}}} \;\)

\(\vec J\): volume current density

In all lines, surface and volume.

\(\vec H \propto \frac{1}{R^2}\)

The magnetic intensity is inversely proportional to the Square of the distance between them.

According to Biot-Savart law,

\(B=\frac{{{\mu }_{0}}}{4\pi }I\frac{dl\times \hat{r}}{{{r}^{2}}}\)

The direction of magnetic field B is the direction of current element dl crossed into the radial direction of the wire , which is in the opposite direction.

The coils are identical and I₁ = I₂

⇒ B₁ = B₂ and hence,

 \(\vec{B}={{\vec{B}}_{1}}+\left( -{{{\vec{B}}}_{2}} \right)=0\)

Concept:

• Magnetic field due to a straight current-carrying conductor: Biot-Savart Law
  • ​Magnetic field B at a radial distance r, due to a wire carrying current is given by:

\(B = \frac{μ_0I}{2π r}\)

Where μ0 is the permeability of free space (4π × 10-7 Tm/A), and I is the current intensity.

Right-hand thumb rule:

According to the right-hand thumb rule, if you curl the fingers of the right hand in the direction of the current, the direction of the thumb gives the direction of the induced magnetic field.

Force on Current-Carrying wire in Magnetic Force

• The magnetic force on a current-carrying wire is given by:

F = I L B Sin θ

Where I is current in the wire, L is the length of the wire, B is the magnetic field and θ is the angle between current and magnetic field.

Explanation:

So, we have the expression

\(B = \frac{μ_0I}{2π r}\)

Here, from the expression, we can see that. the magnetic field is directly proportional to the current in the conductor for a given distance as  μ0  and π are constant. 

So, The strength of the magnetic field will increase at a given point if we increase the magnitude of current in the conductor. This statement is correct. 

Also, The direction can be obtained from right-hand thumb rule is true. 

Now, if the current is the same, we can say that

\(B \propto \frac{1}{r}\)

The field is inversely proportional to the distance from the point. 

So, The strength of the magnetic field will be reduced as we move ahead from the conductor for the same current. So, the statement 'The strength of the magnetic field will be more as we move ahead from the conductor for the same current.' is wrong.  

The next statement to be considered is 

If another current-carrying conductor is placed near a given conductor, it will experience a force. This is because of the magnetic field presence due to the initial current carrying wire. 

So, this statement 'If another current carrying conductor is placed near given conductor, it will experience a force.' is also true.

Concept:

The magnetic field at a distance ‘R’ from a current-carrying conductor carrying current I is given by:

\(B = \frac{{{\mu _0}I}}{{2\pi R}}\)

Application:

Since the two wires are carrying current in the opposite direction, the magnetic field due to wire ‘1’ will be:

\({B_1} = \frac{{{\mu _0}I}}{{2\pi r}}\)

The direction, according the right-hand rule, will be downwards into the plane of the paper/screen.

Magnetic field due to wire ‘2’

Similarly \({B_2} = \frac{{{\mu _0}\left( {2I} \right)}}{{2\pi \left( {3r} \right)}}\)

\(B_2= \frac{{2{\mu _0}I}}{{2\pi 3r}}\) 

The direction for this is also into the plane of the paper.

Adding the two we get:

B = B₁ + B₂

\(= \frac{{{U_0}I}}{{2\pi r}} + \frac{{2{\mu _0}I}}{{2\pi 3r}}\)

\(= \frac{{{u_0}I}}{{2\pi r}}\left[ {1 + \frac{2}{3}} \right]\)