Biot-Savart Law

What is Biot-Savart Law?

Biot-Savart’s law is an equation that gives the magnetic field produced due to a current carrying segment. This segment is taken as a vector quantity known as the current element.

Biot Savart Law

What is the Formula of Biot-Savart’s Law?

Consider a current carrying wire ‘i’ in a specific direction as shown in the above figure. Take a small element of the wire of length ds. The direction of this element is along that of the current so that it forms a vector i ds.

To know the magnetic field produced at a point due to this small element, one can apply Biot-Savart’s Law. Let the position vector of the point in question drawn from the current element be r and the angle between the two be θ. Then,

\(\left | dB \right |=(\frac{\mu _{0}}{4\pi })(\frac{Idlsin\Theta }{r^{2}})\)

Where

  • μ0 is the permeability of free space and is equal to 4π × 10-7 TmA-1.

The direction of the magnetic field is always in a plane perpendicular to the line of element and position vector. It is given by the right-hand thumb rule where the thumb points to the direction of conventional current and the other fingers show the magnetic field’s direction.

Biot Savart Law
In the figure shown above, the direction of the magnetic field is pointing into the page.
This can be expressed in terms of vectors as:
d→B = μ04π i →ds ×^rr2

Let us use this law in an example to calculate the Magnetic field due to a wire carrying current in a loop.

Example of Biot-Savart’s Law

The magnetic field of Current Loop:

Consider a current loop of radius R with a current ‘i’ flowing in it. If we wish to find the electric field at a distance l from the center of the loop due to a small element ds, we can use the Biot-Savart Law as:

d→B = μ04π i d→s ×^rr2

Consider the current element ids at M which is coming out of a plane in the figure. Since r is in the plane of the page, the two of them are perpendicular to each other. Furthermore, the magnetic field produced db is also in the plane of the page.

dB = μ04π i ds . 1. sin 90⁰r2 = μ04π i dsr2

But from the figure,

R2 + l2 = r2

dB = μ04π i dsR2 + l2

Now, if we consider the diametrically opposite element at N, it produces a field such that it’s component perpendicular to the axis of the loop is opposite to that of the field produced at M. Thus only the axial components remain. We can divide the loop into diametrically opposite pairs and apply the same logic.
Also note that from figure that

α = θ

∴ cos θ = R√R2 + l2

Thus,

dB cos θ = μ04π i dsR2 + l2 × R√R2 + l2

The total field will be thus,

B = ∫ μ04π i ds R(R2 + l2)32 = μ04π i R(R2 + l2)32 ∫ ds

B = μ04π i R(R2 + l2)32 × 2πR

B = μ0 i R22(R2 + l2)32

The right hand thumb rule can be used to find the direction of magnetic field.
Related Articles:

  • Derivation of Biot Savart Law
  • Applications of Biot-Savart’s Law

    Some of Biot-Savart’s Law applications are given below.

    • We can use Biot–Savart law to calculate magnetic responses even at the atomic or molecular level.
    • It is also used in aerodynamic theory to calculate the velocity induced by vortex lines.

    What are the importance of Biot-Savart Law?

    Following are the importance of Biot-Savart law:

    • Biot-Savart law is similar to the Coulomb’s law in electrostatics.
    • The law is applicable for very small conductors too which carry current.
    • The law is applicable for symmetrical current distribution.

    Biot-Savart Law example Problems

    Q1. Determine the magnitude of the magnetic field of a wire loop at the center of the circle with radius and current I?

    Ans: The magnitude of the magnetic field of the wire loop is given as:

    \(\frac{\mu _{0}I}{2R}\)

    Q2. A circular coil of radius 5 × 10-2 m and with 40 turns is carrying a current of 0.25 A. Determine the magnetic field of the circular coil at the center.

    Ans: The radius of the circular coil = 5 × 10-2 m

    Number of turns of the circular coil = 40

    Current carried by the circular coil = 0.25 A

    Magnetic field is given as: \(B=\frac{\mu _{0}NI}{2a}\)

    = \(\frac{4\pi \times 10^{-7}T.m/A(40)0.25A }{2.50\times 10^{-2}m}\)

    = 1.2 × 10-4 T

    Q3. Determine the magnetic field at the center of the semicircular piece of wire with radius 0.20 m. The current carried by the semicircular piece of wire is 150 A.

    Ans: The radius of the semicircular piece of wire = 0.20 m

    Current carried by the semicircular piece of wire = 150 A

    Magnetic field is given as: \(B=\frac{\mu _{0}NI}{2a}\)

    The differential form of Biot-Savart law is given as: \(dB=\frac{\mu _{0}I}{4\pi }\frac{dIsin\theta }{r^{2}}\)

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

    \(=\frac{\mu _{0}}{4\pi }\frac{I}{r^{2}}\int dI\)

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

    \(=\frac{\mu _{0}I}{4r}\)

    \(=\frac{4\pi \times 10^{-7}T.m/A(150A)}{4(0.20m)}\)

    =2.4×10-4 T

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