Q. 464.6( 5 Votes )

A charge of 3.14 × 10–6 C is distributed uniformly over a circular ring of radius 20.0 cm The ring rotates about its axis with an angular velocity of 60.0 rad s–1. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the center.

Answer :

Given:


Charge of the ring(q) = 3.14 × 10–6 C


Radius of circular ring(r) = 20 cm = 0.2 m


Angular velocity with which the ring rotates(w) = 60 rad s-1


Diagram:



E1 and E2 denote the electric field due to two point elements 1 and 2 on the circumference of the circular loop. The resultant electric field due to them is E. Similarly, the resultant magnetic field due to the them is B.


Formula used:


Electric field due to a circular loop(E),



where q = charge carried by the ring, x = distance from the center, r = radius of the ring, ϵ0 = electric permittivity of vacuum = 8.85 X 10-12 N m2 C-2


Magnetic field due to a current carrying loop(B),



where μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1, i = current carried by the loop, r = radius of the loop, x = distance from the center



Now, i = q/t, where i = current, q = charge, t = time taken.


Angular velocity(w) = 60 rad/s (given)


Hence, time period of one revolution(T) = 2π/w = 2π/60 s


Hence, current(i) = charge per unit time = q/T = 60q/2π = 30q/π A, where q = charge, T = time period of revolution


Therefore, E/B = (2 x 3.14 × 10–6 x 0.05 x 9 x 109 x 3.14)/ (4 x 3.14 x 10-7 x 30 x 3.14 × 10-6 x 0.22) = 1.88 x 1015 ms-1 (Ans)


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