Question

answer is 200nJ

A charged rod, with Q = 5 nC evenly distributed over its length, is moving along the x-axis with a velocity of u 10 m/s. The rod is in a magnetic potential given by the following equation What is the magnetic potential energy of this system? y [m] x [m]

Answer 1

Given data: Charge (Q) = 5 nc = 5 times 10^-9 c velocity (v) = 10 m/s The given equation v_n rightarrow = y(2 J/Am^2) x + x(2 J/Am^2)y nabla bar v_m = 2 + 2 = 4 J/Am^2 We know that the formula magnetic potential energy (E_n) = (nabla v_m) times Q times v times A rightarrow (1) Substitute nabla v_m = 4 J/Am^2, Q = 5 times 10^-9 c, v = 10 m/s A = 10 m^2 in the above eq No (1), then we will get rightdoublearrow E_m = 4 times 5 times 10^-9 times 10 times 10 = 20 times 10^-7 = 200 times 10^-6 Joules = 200 n J [therefore 1n J = 10^-6 J] therefore magnetic potential energy of this system = 200 nJ