What is the magnitude of the magnetic field 1.5 cm from a long, straight conductor carrying a current of 3.2 A?
1 Answer
Explanation:
It is given that the current carrying wire is long and straight. We assume that it is of infinite length.
The magnitude of magnetic field
#|vecB|=(mu_0I)/(2pir)#
where#mu_0=4pixx10^-7" TmA"^-1# is permeability of free space,#I# is current flowing in the conducting wire and#r# is the radial distance of point of interest.
The direction of the magnetic field which is perpendicular to the direction of current is given by right hand thumb rule as shown in the figure below. Observe that the thumb points in the direction of current and direction in which the fingers curl gives the direction of the magnetic field.
Inserting given values in the expression in SI units we get
