A long, straight wire of length 2 L on the y -axis carries a current I. According to the Biot-Sa

A long, straight wire of length 2 L on the y -axis carries a...

Key Concepts

Biot-Savart Law

The Biot-Savart Law is a fundamental principle in electromagnetism that relates the magnetic field produced by a current element to the geometry of the current and the point at which the field is being calculated. It represents the contribution of infinitesimal current segments to the total magnetic field, integrating these contributions over the current distribution to obtain the final field vector at a given point.

Magnetic Field due to a Current-Carrying Wire

This concept involves computing the magnetic field created by currents flowing through wires. By applying the Biot-Savart Law to a straight wire or segments of it, one can derive expressions that show how the magnetic field depends on factors such as the distance from the wire, the length of the current-carrying segment, and the current itself. For an infinitely long wire, a simplified expression emerges due to the symmetry and simplification of the integration limits.

Integration Techniques in Electromagnetic Problems

Integration is used to sum the contributions of all the differential current elements along the wire. This requires setting up the integral with the correct limits and geometric relationships, often involving trigonometric functions that relate the position along the wire to the observation point. Through careful evaluation of these integrals, one derives the complete expression for the magnetic field at a specific point.

Limit Analysis for Infinite Systems

Limit analysis involves taking parameters, such as the length of the wire, to infinity to simplify the magnetic field expressions. This process exploits the symmetry and uniformity that arise in infinite or very large systems, yielding familiar results such as the magnetic field expression for an infinite straight wire. This approach is practical in many real-world problems where the current distribution can be approximated as infinitely long.

Transcript

00:01 In this question we are given with the expression b of a equals to mu not i over 4 pi integral of sine theta over r squared or d y for the limit minus l to l.

00:12 This expression gives us the value of magnetic field at the point a.

00:17 0 because of current carrying wire of length 2l placed along y -axis.

00:25 So now from this triangle this this angle will be 90 minus theta so in the first case we need to prove that this magnetic field will be equal to mu not i l over 2 pi a square root a square plus l square.

00:51 So first of all we will we can see this angle will be this angle is theta and hence this well angle will be theta minus 90 degree so cosine of this angle which is theta minus 90 degree i'm talking about this one angle this angle will be theta minus 90 degree so will be equal to a over r which is base over hybrid is so this will be a over r and cosine theta minus 90 degree can be written as sine theta or this will be equal to a over r and we will use pythagoras theorem in this triangle and we will get r will be equal to square root of a square plus y square and now by substituting these values in the given expression we will get b of a equal to mu not i over 4 pi integral for the limit minus l to l and sine theta will be replaced by a over r so this will become a over r cube dot dy now we will replace r by square root a square plus y square and we will get b of a equal to mu not i over four pi integral for the limit minus l to l a over this will become a square plus y square to the power three by two dot d y now this limit is from minus l to l this can be converted to zero to l as this is a even function and we will get uh b of a equals to mu not i over twice of this integral will become only here two pi only.

02:48 So this will become the limit for 0 to l and a over a square plus y squared to the power 3 by 2 .d .y.

02:59 Now we will consider this integral as integral i which is a which we will consider it as a indefinite integral and now we will calculate this integral i as an indefinite integral so this will be integral of a over a square plus y square to the power 3 by 2 dot d y in this integral we will substitute y as a tangent phi we cannot consider it as tangent theta as it theta is already a variable in this question so we will substitute y as a tangent 5.

03:43 So we will get dy equal to a secant square phi dot d5.

03:51 Now substituting these values we will get i equal to integral of a dot dy will be replaced by a dot secant square 5 dot d5 over the denominator will be a square plus a square tangent square phi here by simplifying this we will get this integral i as integration of a square second square five dot d phi over here this will be and there was three by two to the power of this bracket so this will become a squared to the power three by two which will be a cube dot this will be one plus tangent square five to the power three and this will be second square 5 to the power 3 by 2.

04:49 So this will become second cube 5.

04:53 By simplifying this we will get i equal to 1 over a integral of 1 over secant 5.

05:01 D5 which is cosine 5.

05:04 D5...

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