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wileyPLUs: HywileyPLUST HelpT Contact Us Cutnell, Physics, 9e College Physics I and II (PHY 205 Assignment Gradebook ent FULL SCREEN PRINTER VERSION BACK Chapter 21, Problem 53 Suppose in the drawing that 12 43.0 A and that the separation between the wires is 0.0268 m. By applying an external magnetic field (created by a source other than the wires) it is possible to cancel the mutual repulsion of the wires. This external field must point along the vertical direction. (a) Does the external field point up or down? (b) What is the magnitude of the external field? Wire 1 Wire 2 Repulsion (a) The external field points (b) Number the tolerance is +/-2% Click if you would like to Show Work for this question: Units Open Show Work Ouestion Attemots: 0 of 5 used SUBMIT ANSWER

Explanation / Answer

magnetic force acting on each wire = F = uo*I1*I2*l/(2*pi*r)

if a magnetic field B acts in down wards direction

the force F1 on wire ,due to B will be opposite to F

F1 = B*I*l

therefore

F1 = F

B*I*l = uo*I1*I2*l/(2*pi*r)

B = uo*I/(2*pi*r)

B = (4*3.14*10^-7*43)/(2*3.14*0.0268)

B = 3.2*10^-4 T

(a) field point down

(b) number = 3.2*10^-4 T

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