The Figure shows two long straight wires perpendicular to the page. The one on t

Question

The Figure shows two long straight wires perpendicular to the page. The one on the right has a current flowing into the page (thus the x symbol) the one of the left has a current out of the page. a) Let us focus first on the wire on the left. What is the direction of the magnetic field due to that wire straight above the wire (to the left of the point marked X)? toward the left towards the right O up direction down direction O into the page O out of the page CHECK b) Still focusing on the wire on the left. What is the direction of the magnetic field due to that wire straight to the left of the wire? toward the left towards the right CO O up direction down direction O into the page O out of the page CHECK Assume that the current in that wire to be 3.40 A, and that the distance separating it from point X is 2.10 m. c) What is in Tesla the magnitude of the the magnetic field due to that wire at pointX? CHECK d) Knowing that the point marked "X" is midway between the two wires, the distance separating the wires is d- 1.40 m and that the current in the wire to the right is 1.90 A, find (in Tesla) the magnitude of the the magnetic field due to both wires at the point marked "X". CHECK e) What is the direction of the magnetic field due to both wires at point "X Express your answer in degrees relative to the conventional positive x-direction. CHECK

Explanation / Answer

a) using the right hand rule for finding the direction of magnetic field ,

magnetic field is towards Right

b)
in the left of left wire ,

the magnetic field is in up direction

c)

Now, magnetic field due to a wire is given as

B = u0*I/(2*pi*d)

B = 4pi *10^-7 * 3.4 /(2 * pi* 2.10)

B = 3.24 *10^-7 T

the magnetic field is 3.24 *10^-7 T

d)

Now, due to right wire ,

at point X ,

B2 = 4pi *10^-7 * 1.90/(2 * pi* 2.10)

B2 = 1.81 *10^-7 T

Now, angle between the magnetic field due to there wire is theta

theta = 180 - 2 * arccos(1.1/(2 * 2.1))

theta = 30.4 degree

Now , resultant is given as

Bnet = sqrt(B^2 + B2^2 + 2 * B1 * B2 * cos(theta)))

Bnet = 10^-7 * sqrt(1.81 ^2 + 3.24^2 + 2 * 3.24 * 1.81 * cos(30.4))

Bnet = 4.89 *10^-7 T

the net magnetic field at X is 4.89 *10^-7 T

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