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Biot Savart Law

Q: Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right- hand rule to find out the direction of the magnetic field inside and outside the loop.

In the following diagram the current is flowing clockwise. If we are applying right hand thumb rule to the left side of the loop then the direction of magnetic field lines inside the loop are going into the table while outside the loop they are coming out of the table.

Q: Derive an expression for the force between two parallel conductors carrying currents. Hence define ampere.

Let the distance between the two conductors carrying current \$I_1\$ and \$I_2\$ be \$d\$.Magnetic field produced by conductor 1 at a distance \$d\$, \$B_1 = \dfrac{\mu_o}{4\pi}\dfrac{2I_1}{d}\$Force acting the conductor 2 placed in this magnetic field \$F =B_1 I_2 L\$Thus we get \$F =\dfrac{\mu_o}{4\pi} \dfrac{2I_1 I_2}{d} L\$Force er unit length of conductor 1 \$f =\dfrac{\mu_o}{4\pi} \dfrac{2I_1I_2}{d}\$Now put \$d = 1 m\$, \$I_1=I_2=1 A\$ and \$L = 1 m\$We get \$F = \dfrac{\mu_2}{2\pi} = 2\times 10^{-7}\$ \$N\$Thus we define ampere as the current flowing in each conductor separated by a unit distance so that one conductor applies a force of \$2\times 10^{-7}\$ N on a unit length of another parallel conductor.

Q: A long straight wire of radius a carries a steady current. The current is uniformly distributed across its cross-section.the ratio of the magnetic field at \$\frac{a}{4}\$ inside and \$\frac{a}{4}\$ outside from the surface of wire is:

Solution : We know,magnetic field B in inside points = \$\dfrac{μ_0ir}{2πR^{2}}\$ [where,R = radius,r = distance of the inside point from the axis and i = total current]Distance from the axis at \$\dfrac{a}{4}\$ inside = \$(a-\dfrac{a}{4})\$ =\$\dfrac{3a}{4}\$So,Magnetic field at \$\dfrac{a}{4}\$ inside = \$\dfrac{μ_0×i×\dfrac{3a}{4}}{2πa^{2}}\$= \$\dfrac{3μ_0i}{8πa}\$Distance from the axis at \$\dfrac{a}{4}\$ outside = \$(a+\dfrac{a}{4})\$We also know,magnetic field B at outside points = \$\dfrac{μ_0i}{2πr}\$ [r=distance of the outside point from the axis]So,magnetic field at \$\dfrac{a}{4}\$ outside = \$\dfrac{μ_0i}{2π×\dfrac{5a}{4}}\$= \$\dfrac{2μ_0i}{5πa}\$The ratio of the magnetic field at \$\dfrac{a}{4}\$ inside and \$\dfrac{a}{4}\$ outside the surface of the wire = \$\dfrac{\dfrac{3μ_0i}{8πa}}{\dfrac{2μ_0i}{5πa}}\$=15:16So,the correct option is C.

Q: The Biot Savart's Law in vector form is

The Biot Savart's Law we know is given by:\$\vec{\delta B} = \dfrac{\mu_0}{4\pi}\dfrac{I dl \sin(\theta)}{r^2}\hat{r}\$, where \$\theta\$ is the angle between the line element \$\vec{dl}\$ and the radial unit vector \$\hat{r}\$. Now we know: \$\hat{r} = \dfrac{\vec{r}}{r}\$, using this and the cross product of two vectors we get; \$\vec{\delta B} = \dfrac{\mu_0}{4\pi}\dfrac{I(\vec{dl}\times \vec{r})}{r^3}\$

Q: A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in the given Figure. The direction of magnetic field at a point will be North to South

The direction of the magnetic field is perpendicular to the wire and is in the direction the fingers of your right hand would curl if you wrapped them around the wire with your thumb in the direction of the current.So in this case, the direction of magnetic field from north to south is at point located directly below the wire.

Q: State and explain Biot-Savart Law.

Hint: The Biot-Savart law states how the value of the magnetic field at a specific point in space from one short segment of current-carrying conductor depends on each factor that influences the field.Solution:Step1: Application of Biot-Savart lawThe Biot-Savart Law is used to determine the magnetic field intensity B near a current carrying conductor in terms of magnetic field intensity generated by its source current element. Consider a wire carrying an electric current I and also consider an infinitely small length of a wire dl at a distance r from point AStep2: Statement of Biot-Savart lawIt is clear from above free body diagram that, Biot-Savart Law states that magnetic intensity dB at a point A due to current I flowing through a small element dl is,a) Directly proportional to current (I). b) Directly proportional to the length of the element (dl). c) Directly proportional to the sine of angle θ between the direction of current and the line joining the element dl from point A. d) Inversely proportional to the square of the distance (x) of point A from the element dl. \$dB\alpha \dfrac{{Idl\sin \theta }}{{{r^2}}}\$\$dB = K\dfrac{{Idl\sin \theta }}{{{r^2}}}\$ Where, \$K = \dfrac{{{\mu _0}{\mu _r}}}{{4\pi }}\$ μ0= Absolute permittivity of air in vacuum μr= relative permittivity.

Q: \$AB\$ is a current carrying conductor in the plane of the paper as shown in figure. What are the directions of magnetic fields produced by it at points \$P\$ and \$Q\$?

Applying the right hand thumb rule, the direction of magnetic field would be anti-clockwise as shown in diagram around the direction of current. So, the magnetic field at point P would be into the plane of paper. At point Q, the direction of magnetic current would be out from the plane of paper. The strength of the magnetic field is stronger near the conductor and weaker as we move away from the conductor as the strength of the conductor is inversely proportional to the distance from the conductor. Given that the distance of point P from the conductor is greater than the distance of point Q to the conductor, the magnetic field would be stronger near point Q than near point P.

Q: State Biot-Savart law.

Bio Savart Law: The magnetic field at any point due to an element of a conductor carrying current is(1) directly proportional to (a) the strength of the current \$i\$,(b) length of the element \$dl\$ (c) sine of the angle \$\theta\$ between the element in the direction of current and the line joining the element to the point P.i.e., \$dB \propto i, dB \propto \delta, dB \propto sin \theta\$(2) inversely proportional to the square of the distance r of the point P from the centre of the element.\$dB = \dfrac{\mu_0}{4 \pi} \times \dfrac{I dl sin \theta}{r^2}\$

Q: A current flows in a conductor from east to west. The direction of the magnetic field at a point above the conductor is:

By right-hand rule, if the right-hand thumb points in the direction of a current, the magnetic field at a point will be given by the direction of curl of fingers at that point. Hence, in this case, as the thumb points from east to west the fingers at a point above will point in the north direction. Hence, a direction of a magnetic field will be towards the north.

Q: Two parallel long wires carry currents \$i_{1}\$ and \$i_{2}\$ (\$i_{1}\$ > \$i_{2}\$). When the currents are in the same direction, the magnetic induction at a point midway between the wires is \$10\ \mu T\$. If the direction of \$i_{2}\$ is reversed, the induction becomes \$30\ \mu T\$. The ratio of \$\dfrac{i_{1}}{i_{2}}\$is :

1st case: \$B=B_1-B_2\$\$10\times10^{-3}=\dfrac{\mu_0i_1}{2\pi r}-\dfrac{\mu_0i_2}{2\pi r}\$\$\dfrac{\mu_0}{2\pi r}(i_1-i_2)=0.01\$2nd case:\$B=B_1+B_2\$\$30\times10^{-3}=\dfrac{\mu_0i_1}{2\pi r}+\dfrac{\mu_0i_2}{2\pi r}\$\$\dfrac{\mu_0}{2\pi r}(i_1+i_2)=0.03\$\$\therefore\dfrac{i_1-i_2}{i_1+i_2}=3\$\$i_1-i_2=3i_1+3i_2\$\$\dfrac{i_1}{i_2}=2\$

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Important Questions

Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right- hand rule to find out the direction of the magnetic field inside and outside the loop.

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Derive an expression for the force between two parallel conductors carrying currents. Hence define ampere.

Medium

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A long straight wire of radius a carries a steady current. The current is uniformly distributed across its cross-section.the ratio of the magnetic field at $\frac{a}{4}$ inside and $\frac{a}{4}$ outside from the surface of wire is:

Easy

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The Biot Savart's Law in vector form is

Medium

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A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in the given Figure. The direction of magnetic field at a point will be North to South

Medium

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State and explain Biot-Savart Law.

Medium

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$A B$ is a current carrying conductor in the plane of the paper as shown in figure. What are the directions of magnetic fields produced by it at points $P$ and $Q$ ?

Medium

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State Biot-Savart law.

Medium

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A current flows in a conductor from east to west. The direction of the magnetic field at a point above the conductor is:

Medium

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Two parallel long wires carry currents $i_{1}$ and $i_{2}$ ( $i_{1}$ > $i_{2}$ ). When the currents are in the same direction, the magnetic induction at a point midway between the wires is $10 μ T$ . If the direction of $i_{2}$ is reversed, the induction becomes $30 μ T$ . The ratio of $\frac{i _{1}}{i _{2}}$ is :

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Biot Savart Law

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REVISE WITH CONCEPTS

Magnetic Field Due to Current Carrying Element - Biot - Savart Law

QUICK SUMMARY WITH STORIES

Magnetic Field Due to Current Carrying Element - Biot-Savart Law

Biot-Savart Law -Magnetic field Strength due to a Current Element.

Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right- hand rule to find out the direction of the magnetic field inside and outside the loop.

Derive an expression for the force between two parallel conductors carrying currents. Hence define ampere.

A long straight wire of radius a carries a steady current. The current is uniformly distributed across its cross-section.the ratio of the magnetic field at 4a​ inside and 4a​ outside from the surface of wire is:

The Biot Savart's Law in vector form is

A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in the given Figure. The direction of magnetic field at a point will be North to South

State and explain Biot-Savart Law.

AB is a current carrying conductor in the plane of the paper as shown in figure. What are the directions of magnetic fields produced by it at points P and Q?

State Biot-Savart law.

A current flows in a conductor from east to west. The direction of the magnetic field at a point above the conductor is:

Two parallel long wires carry currents i1​ and i2​ (i1​ > i2​). When the currents are in the same direction, the magnetic induction at a point midway between the wires is 10 μT. If the direction of i2​ is reversed, the induction becomes 30 μT. The ratio of i2​i1​​is :

A long straight wire of radius a carries a steady current. The current is uniformly distributed across its cross-section.the ratio of the magnetic field at $\frac{a}{4}$ inside and $\frac{a}{4}$ outside from the surface of wire is:

$A B$ is a current carrying conductor in the plane of the paper as shown in figure. What are the directions of magnetic fields produced by it at points $P$ and $Q$ ?