Two infinitely long straight wires A1 and A2 carrying currents I and 2I flowing in the same directions are kept 'd' distance apart. Where should be the third straight wire A3 be placed b/w A1 and A2 so that it experiences no net force due to A1 and A2? Does the net force on A3 depends on the current flowing through it??
Answers
Here, let the third wire is placed at distance of x from first wire .
So, magnetic field at distance x due to first wire, B = (μI)/(2πx)
Thus, force on ΔL length of wire A3 = I1 * ΔL * B
= I1 * ΔL * (μI)/(2πx)
where, μ = 4π * 10^-7 T.m/A
I = current in wire A1
I1 = current in wire A3
As, first and third wires (A1 and A3) carry current in same direction , they attract each other .
Also, second and third wires (A2 and A3) carry current in same direction , they attract each other .
So, magnetic field at distance d-x due to second wire, B = (μ*2I)/[2π(d-x)]
Thus, force on ΔL length of wire A3 = I1 * ΔL * B
= I1 * ΔL * (μ*2I)/[2π(d-x)]
Here, as there is no net force on wire A3 , these two forces cancel each other .
=> I1 * ΔL * (μI)/(2πx) = I1 * ΔL * (μ*2I)/[2π(d-x)]
=> (1)/(x) = (2)/[(d-x)]
=> d-x = 2x
=> 3x = d
=> x = d/3
Thus, the third straight wire A3 be placed at distance of d/3 from wire A1 to its right .
Also, the net force on A3 does not depends on the current flowing through it as it becomes to zero .
Answer:
if at a place of A1 A2 A3 a,b, c is given. so we will use this solution
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