Question

(d
39 + 512
7n, +3
5 +7
sn +3n.
A copper rod of length 0.19 m is moving parallel to a lon
wire with a uniform velocity of 10 ms. The long wir
carnes 3 A current and is perpendicular to the rod. The end
of the rod are at distances 0.01 m and 0.2 m from the wir
The emf induced in the rod will be
(a) 100V ( 20uV (c) 30 V (d) 40 V​

Answer 1

Answer:

Explanation:

As shown in Fig.3.55, consider an element of length dy at a distance y from the wire, then at this position of the element, the magnetic field due to the current - carrying wirePQ will be B=μ04π2Iy into the plane of the paper.

So, the emf induced in the element dε=Bvdy=μ04π2Iyvdy and hence the emf induced across the ends of the rod due to its motion in the field of the wire,

ε=∫badε=μ04π2Iv∫badyy, i.e., ε=μ04π2Ivloge(ba)

Substituting the given data with b=(a+l),

ε=10−7×2×5×10loge=0.200.01=10−5×loge20

=30μV