QuestionDetails: Two striaght wire carrying conventional current I areconnceted

Question

QuestionDetails: Two striaght wire carrying conventional current I areconnceted by a three-quarter arc of radius R1 and aone-quarter-circular arc of radius R2. Electric fields are alsopresent in this region, due to charges to charges not shown on thedrawing.    Two arcs and two straight wires An electron is moving down seth speed v as it passes throughthe center C of the arc, and at that instant the net electric andmagnetic force on the electron is zero. (the Earth's mangetic fieldis negligible.) (a) what is he direction of he magnetic field at the center C,due to the current I? Explain briefly. (b) what is the dirction of the electric field at the centerC? Explain your reasoning clearly. (c) what is he magnitude of the magnetic field at at thecenter C, due to the current I? (d) what is the mangitude of the electric field at the centerC? Explain briefly.

Explanation / Answer

(a)The direction of the magnetic field at the cente C,due tothe current I is obtained by Ampere's right hand thumbrule.According to this rule,when the thumb of the right hand pointsin the direction of the current flow then the curling of fingers ofthe hand indicate the direction of the magnetic.As the current inthe wire is carried in the upward direction therefore the magneticfield direction is into the plane of the board. (b)The electric field at the center C is due to the movingelectron.The electron is a negatively charged particle thereforethe lines of force due to the electron converge on the electron.Asthe lines of force indicate the direction of electricfield,therefore the electric field at the center C converges at thecenter. (c)the magnetic field due to a circle of radius R1is B = (oI/2R1) o = 4 * 10-7N/A2 As the circle is now three-quarter arc of radius R1therefore we get (2R1 - (R1/2)) =(3R1/2) or B1 =(oI/(3R1/2)) =(2oI/3R1) (d)The magnitude of the electric field at the center Cis E = k * (q/R12) ----------(1) k = (1/4o) = 9 * 109Nm2/C2 and q = -1.6 * 10-19C we know that the electric field due to a charge at point pwhich is at a distance r from it is E = k * (q/r2) N/C therefore,the electric field at the center due to the electronis given by equation (1)

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