Straight Conductor Current Assignment Help

Electromagnetism - Straight Conductor Current

Straight Conductor Current

Consider a straight conductor XY lying in the plane of paper carrying current I in the direction X to Y let P be a point at a perpendicular distance a from the straight conductor. Clearly, PC = a let the conductor be made of small current elements consider a small current element IdI of the straight conductor at O. let  be the position vector of P w.r.I current element and θ be the angle between I dl‾ and r and CO=I.

According to Biot-savart law the magnetic field induction (magnetic flux density) at point p due to current element IDI is given by 

db‾  = μ0/ 4π I dI X r / r

Odb‾ = μ0 / 4π I dl‾  sin θ / r2

In rt. Angled ? POC θ + ∅ = 90° 

Or θ = 90° -∅

And sin ∅ = sin (90° = ∅) = cos ∅

Also cos ∅ = a /r or r = a / cos ∅ 

And tan ∅ = I /a or I = a tan ∅

Differentiating it we get

dI = a sec2 ∅ d ∅

Putting the values in (5) from (6) (7) and (8) we get

dB = μ0 / 4 π I ( a sec2 ∅ d ∅ ) cos ∅ / (a2 / cos2 ∅)

= μ0 / 4 π I/a cos ∅d∅

The direction of dB according to right hand thumb rule will be perpendicular to the plane of paper and directed inwards. As all the current elements of the conductor will also produce magnetic field in the same direction therefore the total magnetic field at point p due to current through the whole straight conductor XY can be obtained by integrating Eq. (9) within the limits - ∅1 and ∅2 thus

B = ∫dB = μo / 4 π I/a∫ cos ∅d ∅ - ∅1 -∅1 = μo / 4 π I/a [sin ∅] - ∅1 = μo / 4 π I/a( sin ∅1 + sin ∅2)

Special cases when the conductor XY is of indented length and the point p lies near the centre of the conductor then ∅1 = ∅2 = 90°

So, BI = μ0 / 4 πa {sin 90° + sin 90°] = μ0 / 4 π 2 I / a 

(ii) When the conductor XY is of infinite length but the point p lies near the end y (or X)

Then 1 = 90° and ∅2 = 0° 

So B = μ0 / 4 πa {sin 90° + sin θ) = μ0I / 4 πa

Thus magnetic field due to an infinite long linear conductor carrying near its centre is twice than that near its one of the ends 

(iii) if length of conductor is finite say L and point P lies on right bisector of conductor, then ∅= ∅2 = ∅ and

Sin ∅ L / 2 / a2 + (L/2)2 = L / 4a2 + L

Then B = μ0/4π 2 I / a sin ∅ = μ0/4π 2 I / a L / 4a2 + L2

Direction of magnetic field the magnetic field lines due to straight conductor carrying current are in the form of concentric circles with the conductor as centre lying in a place perpendicular to the straight conductor. The direction of magnetic field lines is anticlockwise if the current flows from A to B in the straight conductor and is clockwise if the current flows from B to A in the straight conductor the direction of magnetic field lines is given by right hand thumb rule or Maxwell cork screw rule. 

Right hand thumb rule according to this rule if we imagine the linear conductor to be held in the gap of the right hand so that the thumb points in the direction of current then the curvature of the fingers around the conductor will represent the direction of magnetic field lines 

Maxwell’s cork screw rule: According this 4rule if we imagine a right handed screw paved along the current carrying linear conductor be rotated such that the screw moves in the direction of flow of current then the direction of rotation of the thumb gives the direction of magnetic lines of force. - Straight Conductor Current Assignment Help, Straight Conductor Current Homework Help, Straight Conductor Current Assignment Tutors, Straight Conductor Current Solutions, Straight Conductor Current Answers, Electromagnetism Assignment Tutors

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