A very long straight wire carries a current I. A plane rectangular coil of high
ID: 2265003 • Letter: A
Question
A very long straight wire carries a current I. A plane rectangular coil of high resistance, with sides
of length a and b, is coplanar with the wire. One of the sides of length a is parallel to the wire and a distance D from it; the opposite side is further from the wire. The coil is moving at a speed v in its own plane and away from the wire.
I have included my workings/thoughts.
(a) Find the e.m.f. induced in the coil.
I know that i first have to calculate the magnetic field of the wire: B(y)=??I/2?y
Then the e.m.f, ?=-d/dt(?(B))= -d/dt ?B?ds = -d/dt(a.?B.dy) with upper limit of integral=D+b and lower limit of integral D (*)
I have been given the answer of ?=[??Ivab]/[2?D(D+b)] (**)
What i am having trouble with is the intermediate step getting from (*) to (**)
(b) Let R be the resistance of the coil. Calculate the force needed to move the coil with speed v as described, and show that the mechanical power used to move it is equal to the rate at which heat is generated in the coil.
I know that Pmech=F.v and that Pheat= v
Explanation / Answer
B = ((mu0)*I)/(2*pi*r)
emf = (((mu*I)/(2*pi*D))*a*v)+(((mu*I)/(2*pi*(b-D)))*a*v)
= (mu*I)(a*v*(b-D)+a*v*D)/(2*pi*D*(b-D))
= (mu*I*a*v*b)/(2*pi*D*(b-D))
The answer provided is:
emf = (mu*I*a*v*b)/(2*pi*D*(D+b))