iPad 6:19 PM 49%. \" Problem 28.5 MasteringPhysics Three long parallel wires are
ID: 1452915 • Letter: I
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iPad 6:19 PM 49%. " Problem 28.5 MasteringPhysics Three long parallel wires are 3.5 cm from one another. (Looking along them, they are at three corners of an equilateral triangle.) The current in each wire is 8.25 A,but its direction in wire M is opposite to that in wires N and P (see the figure). N3.5 cm P Q Tap image to zoom Part A Determine the magnitude of magnetic force per unit length on wire M due to the other two. Express your answer to two significant figures and include the appropriate units. Fu= 1 Value Units Submit Give Up Part B Determine the direction of magnetic force per unit length on wire M due to the other two.Explanation / Answer
Current through each wire is same and distance between them also same.
hence magnitude of force on one wire due to another will be same
F / L = u0 I^2 / 2 pi d
= (4 x pi x 10^-7 x 8.25^2 ) / (2 x pi x 0.035 )
= 3.89 x 10^-4 N/m
when direction of current is same in both wires then they attarct otherwise repel.
A) N and P both will repel M along there joining line.
and each will make 30 deg with vertical.
Fnet = F/L(sin30i + cos30j ) + F/L (-sin30i + cos30j)
Fnet = 6.74 x 10^-4 N/m
b) direction- upward
c) due to P it will be along x axis.
due to M , it will be 60 deg below -ve x axis.
Fnet = B/L (i) + B/L (-cos60i - sin60j)
Fnet = (1.945i - 3.37j) x 10^-4 N/m
magnitude = sqrt(1.945^2 + 3.37^2) = 3.89 x 10^-4 N/m
d) direction =360 - tan^-1(3.37 / 1.945) = 300 deg
e) due to N it will be along -ve x axis.
due to M , it will be 60 deg below x axis.
Fnet = B/L (-i) + B/L (cos60i - sin60j)
Fnet = (-1.945i - 3.37j) x 10^-4 N/m
magnitude = sqrt(1.945^2 + 3.37^2) = 3.89 x 10^-4 N/m
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