CBSE Class 12-science Answered
pl derive the expression for the field due to straight wire using Biot Sawart's law
Asked by Rajeev | 21 Feb, 2016, 11:29: PM
Expert Answer
Consider a straight wire 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 wire conductor. Clearly, PC = a. Let the conductor be made of small current elements. Consider a small current element of the straight wire conductor at O. Let be the position vector of P w.r.t. current element and θ be the angle between and . Let CO = .
According to Biot-Savart's law, the magnetic field (i.e., magnetic flux density or magnetic induction) at point P due to current element is given by
or
In right angled ΔPOC, θ + Φ = 90º or θ = 90º-Φ
Therefore, sin θ = sin (90º-Φ) = cos Φ ... (2)
Also,
And,
Differentiating it we get,
d = a sec2ΦdΦ ... (4)
Putting the values of eqn (2), (3) and (4) in eqn (1), we get,
The direction of , according to the 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 the current through the whole straight wire conductor XY can be obtained by integrating equation (5) within the limits -Φ1 and +Φ2.
Thus,
Answered by Faiza Lambe | 22 Feb, 2016, 02:56: PM
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