What current is needed to generate the magnetic field strength of 5.0×10−5T at a point 1.5 cm from a long, straight wire? Expres
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Answer:
Explanation:
Since the wires attract each other , the direction of current will be same in both the wires .
Let I be current in wire which is along x - axis
force of attraction per unit length between the two current carrying wire is given by
x
where I₁ and I₂ are currents in the wires and d is distance between the two
Putting the given values
285 x 10⁻⁶ = 10⁻⁷ x
I₂ = 16.76 A
Current in the wire along x axis is 16.76 A
To find point where magnetic field is zero due the these wires
The point will lie between the two wires as current is in the same direction.
Let at y = y , the neutral point lies
k 2 x = k 2 x
25.5y = 16.76 x .3 - 16.76y
42.26 y = 5.028
y = .119
= .12 m
Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.
