Deduce an expression for the gravitational potential
1 answer:
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Answer:
Explanation:
we can consider an element of radius r < a and thickness dr. and Area of this element is
since current density is given
then , current through this element will be,
integrating on both sides between the appropriate limits,
Magnetic field can be found by using Ampere's law
for points inside the wire ( r<a)
now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.
by applying the Ampere's law, we can write
by symmetry will be of uniform magnitude on this loop and it's direction will be tangential to the loop.
Hence,
now using equation 1, putting the value of k,
B)
now, for points outside the wire ( r>a)
consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.
applying the Ampere's law
by symmetry will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence
again using,equaiton 1,