Set 54 Problem number 1

Problem

A current of 7.6 Amps flows in a straight wire segment of length .058 meters. Find the strength of the magnetic field at a point lying 6.6 meters from the segment, provided that the vector from the segment to the point is perpendicular to the segment.

Solution

The magnetic field at a point P due to a short current segment, with point P lying 'at a perpendicular' to the segment, is proportional to the current and the length of the segment.

Here a segment is 'short' if the field is measured at a distance large with respect to the length of the segment.
P lies nearly 'at a perpendicular' to the segment if a vector from the segment to P is at nearly a right angle.

The specific value of the magnetic field is given by B=k ' (IL)/r ^ 2.

In this situation we have I = 7.6 Amps, L= .058 m, and r= 6.6 meters.

Thus

IL= .4408 Amp meters, and
B = .0000001 Tesla / Amp meter)( .4408 Amp m)/( 6.6 m) ^ 2 = 1.011E-09 Tesla.

Generalized Solution

Just as the electric field strength E = k q / r^2 of a point charge q falls off as the inverse square of distance from the source q, the magnetic field B = k' (IL) / r^2 of a short charge-and-length segment IL falls of as the inverse square of its distance from the source IL.

We can consider IL to be the source of the magnetic field, just as q is the source of electric field.

The magnetic field is a little trickier than the electric field, since it depends not only on distance but on the orientation of the point with respect to the direction of the current I from which it arises. However

if the orientation is such that a line from the source IL to the point is perpendicular to I, the equation B = k' (IL) / r^2 holds.

The direction of the magnetic field is found by the right-hand rule:

if the fingers of the right hand are directed in the direction of the current, and oriented so that when the fingers are current they move toward the direction of the line from source to point, the thumb points in the direction of the magnetic field.