Two long straight wires pass through the middle of the picture. Wire 1 carries a

Question

Two long straight wires pass through the middle of the picture. Wire 1 carries a current up, while wire 2 (perpendicular to wire 1) carries a current to the right. Four points are shown, labelled P, Q, R, and S. The ratio of the current in wire 1 to the current in wire 2 is 2. Defining out of the screen to be positive for magnetic field, we can say that the net field at point S is -B (negative, because it is into the screen). (a) In terms of B, what is the net magnetic field at point P? (Do not include the B in the answer box below.) Note that this is equivalent to asking for the ratio of the field at P to the field at S. B (b) In terms of B, what is the net magnetic field at point R? (Do not include the B in the answer box below.) B (c) Now, let's figure out what B represents, in units of teslas. Note that this answer should be positive. If the current in wire 2 is 3 A, and each small square in the picture measures 5 cm times 5 cm, what is B? T

Explanation / Answer

let each box has width d.

a) at point s,

-B = mue*I1/(2*pi*d) + mue*I2/(2*pi*d)

-B = 3*mue*I2/(2*pi*d)

B = -3*mue*I2/(2*pi*d) (since I1 = 2*I2)

at point P,

magnetic field = mue*I1/(2*pi*d) - mue*I2/(2*pi*2*d)

= mue*2*I2/(2*pi*d) - mue*I2/(2*pi*2*d)

= (3/2)*mue*I2/(2*pi*d)

= -B/2

= -0.5 B <<<<<<<<--------------Answer

b) at point R

magnetic field = mue*I2/(2*pi*2*d) - mue*I1/(2*pi*3*d)

= mue*I2/(2*pi*2*d) - mue*2*I2/(2*pi*3*d)

= -0.167*mue*I2/(2*pi*d)

= 0.167 B <<<<<<<<--------------Answer

c)

B = 3*mue*I2/(2*pi*d)

= 3*4*pi*10^-7*3/(2*pi*5*10^-2)

= 3.6*10^-5 T <<<<<<<<--------------Answer

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