Two parallel vertical current carrying wires both have length l = 69.2 cm. are s
ID: 1610485 • Letter: T
Question
Two parallel vertical current carrying wires both have length l = 69.2 cm. are spaced d = 4.62 cm apart, and experience a repulsive force of magnitude 1.35 mu N. If the current in wire 1 is I_I = 0.379 A, what is I_2? a) 2.38 A up b) 0.379 A down c) 0.595 A up d) 1.19 A down Which integral below correctly gives the electric field strength E at point P on the perpendicular axis of a line of charge of length 2a and uniform linear charge density lambda, at a distance x from the line? a) E = integral^a_-a 1/2 pi epsilon_0 lambda y dy/(x^2 + y^2)^2 b) E = integral^a_-a 1/4 pi epsilon_0 lambda y dy/(x^2 + y^2) c) E = integral^a_-a 1/4 pi epsilon_0 lambda y dy/(x^2 + y^2)^3/2 b) E = integral^a_-a 1/4 pi epsilon_0 lambda y dy/(x^2 + y^2)^3/2 A current loop is in a uniform magnetic field. Which describes the rotation of the loop? a) The loop tends to rotate so that the direction of the normal vector is in the same direction of the magnetic field vector. b) The loop tends to rotate so that the direction of the normal vector is perpendicular to the direction of the magnetic field vector. c) The loop tends to rotate so that the current is parallel to the direction of the magnetic field vector. d) The loop tends to rotate so that the net force vector on the loop is maximized.Explanation / Answer
27. Taking a small element of length dy as shown in the figure, it's charge dq = lambda*dy
The electric field due to it at the point P consists of two components. One pointing rightwards and one pointing downwards.
Now, first we need to appreciate that the net electric field of the vertical direction cancels out and is zero.
The net rightwards electric field is the Summation of all the small components contribution.
dE (rightwards direction component)= (1/4€) * dq* cos(theta) / (x2 + y2) = (1/4€) * x * lambda*dy / (x2 + y2)3/2
Integrating the above equation over dy from -a to a
Answer is option D
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