Rank the strength(magnitude) of the total magnetic field at point P. /42/mod res

Question

Rank the strength(magnitude) of the total magnetic field at point P. /42/mod resource/content/1/Review.p hown below are six situations where three long straight parallel wires carry currents either into or o e page. In each situation, point P is midway between two adjacent wires 2A 2A A 2A 2 3 A 2 A P 2A 3 A pxd 3 A 2 A Kank the strength (magnitude) of the total magnetic field at point Greatest 1 OR, the total magnetic field at P has the same (but not zero) strength for all six of these situations. OR, the total magnetic field is zero at P for all six of these situations. OR, the ranking for the magnetic fieids cafnot be det Carefully esplain your reasoning ermined.

Explanation / Answer

magnetic field due to infinite long wire = B = uo*I/2*pir

uo = 4*pi*10^-7

uo/pi = 4*10^-7

A)

Bnet = -uo*2/(2*pi*d/2) - uo*2/(2*pi*d/2) - uo*3/(2*pi*3d/2)

Bnet = (uo/pi*d)*( -2 - 2 - 1) = -5*uo/pid

Bnet = -20*10^-7/d   T

B)

Bnet = +uo*2/(2*pi*d/2) + uo*2/(2*pi*d/2) - uo*3/(2*pi*3d/2)

Bnet = (uo/pi*d)*( +2 + 2 - 1) = 3*uo/pid

Bnet = 12*10^-7/d   T

C)

Bnet = -uo*2/(2*pi*d/2) + uo*2/(2*pi*d/2) - uo*3/(2*pi*3d/2)

Bnet = (uo/pi*d)*( -2 + 2 - 1) = -1*uo/pid

Bnet = -4*10^-7/d   T

D)

Bnet = -uo*2/(2*pi*d/2) + uo*3/(2*pi*d/2) + uo*3/(2*pi*3d/2)

Bnet = (uo/pi*d)*( -2 +3 + 1) = 2*uo/pid

Bnet = 8*10^-7/d   T

E)

Bnet = -uo*2/(2*pi*d/2) - uo*2/(2*pi*d/2) + uo*3/(2*pi*3d/2)

Bnet = (uo/pi*d)*( -2 - 2 + 1) = -3*uo/pid

Bnet = 12*10^-7/d   T

F)

Bnet = uo*3/(2*pi*d/2) - uo*3/(2*pi*d/2) - uo*3/(2*pi*3d/2)

Bnet = (uo/pi*d)*( 3 - 3 - 1) = -1*uo/pid

Bnet = -4*10^-7/d   T

B1 = 20*10^-7

B2 = 12*10^-7

B3 = 4*10^-7

B4 = 8*10^-7

B5 = 12*10^-7

B6 = 4*10^-7

1 > 2 = 5 > 4 > 3 = 6

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