The drawing shows an end-on view of three wires. They are long, straight, and perpendicular to the plane of the paper. Their cross-sections lie at the corners of a square. The currents in wires 1 and 2 are 1 = 2 = and are directed into the paper. What is the direction of the current in wire 3, and what is the ratio 3⁄, so that the net magnetic field at the empty corner is zero?

l3/l = 2

Explanation:

Suppose B1 is the magnetic field due to cable 1 and B2 is the field due to cable 2. Therefore, the magnitude of the field at a distance from the cable is equal to:

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

B1-2 = (B1^2 + B2^2)^1/2 = uo*I)/(2*pi*r)^2 + (uo*I)/(2*pi*r)^2^1/2 = (2^1/2)*(uo*l)/(2*pi*r)

From this equation we can say that the direction of the field will be from the corner that is empty and cable 3:

B3 = B1-2

(uo*l3)/(2*pi*(2^1/2)*r = (2^1/2)*(uo*l)/(2*pi*r)

From here we have that the relationship between both currents will be equal to:

l3/l = 2

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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