Acylindrical conductor with a circular cross section has a radius a and a resistivity p and carries a constant current i. (take the current to be coming out of the page when the cross-sectional view of the conductor is in the plane of the page.) for the following questions, express your answer in terms of the variables i, a,p, and appropriate constants (u0,e0, and pi).a) what is the magnitude of the electric field vector e at a point just inside the wire at a distance a from the axis? b) what is the magnitude of the magnetic field vector b at the same point? c) what is the magnitude of the poynting vector s at the same point? d) what are the directions of these three vectors?

Explanation:

a)

Using Ohms Law

Where J is the current density

and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor just inside at a radius a (and anywhere else).

b)

Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current

such that

and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.

c)

The Poynting vector is given by

and by the right hand rule it’s always pointing in towards the center of the conductor.

d)

Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.