The law of Biot and savart

Solution Preview

2) A current I flows along a wire that makes a right angle bend, as shown in the figure. If this right angle bend lies at the origin and the wire carrying the incoming current lies on the negative y-axis (and extends to large negative distances), then the part of the wire carrying the outgoing current lies on the positive x axis (and extends to large positive distances). Use the law of Biot and savart to find the magnetic field at the point P (a,-a)in the x-y plane. Then find the magnetic field at the point Q whose coordinates are (2a, -a).

Y
I X

. P . Q (2a, -a)
(a, -a)

I

Solution:

Idl r
P

As per Biot-Savart law, magnetic flux density dB at a point P due to a current element Idl is given by :

dB = (μ0/4Π) Idl X ř/r2 (X indicates cross product) where r is the position vector of point P with respect to the current element and ř is the unit vector of r.

Magnitude of vector dB is given by : dB = (μ0/4Π) Idl sinθ/r2 (where Ɵ is the angle between the vectors Idl and r)

Direction of dB is given by the right hand rule of vector multiplication according to which vector dB is perpendicular to the plane in which vectors Idl and r lie. The direction (left or right) is determined as follows : curl the fingers of the right hand such that they point from the first vector (i.e. Idl) towards the second vector (r), then the direction in which the stretched out thumb points gives the direction of dB.
Application of Biot-Savart law to a straight current carrying conductor

Ɵ
IdL
r φ1
...