Question

A stiff wire 44.5 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y=2x in the xy plane. A current of 24.0 A flows in the wire-down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field given by B =(0.318i)T.

A) Determine the magnitude and direction of the total force on the wire. (The direction must be theta below the negative y-axis).

Answer 1

Magnetic force due to current carrying wire is given by:

F = i*LxB

For wire, which is downward in the z-axis

i = current in wire = 24 A

L = length of wire = 44.5/2 = 22.25 cm = 0.2225 m

Since wire is downward in z-axis, So

So magnetic force on this wire will be

F1 = 24.0*(-0.2225 k)x(0.318 i)

F1 = -24.0*0.2225*0.318*(kxi)

F1 = -1.698 j

For wire which is along the line = y = 2x in the xy plane

i = current in wire = 24 A

L = length of wire = 44.5/2 = 22.25 cm = 0.2225 m

Since wire is in xy plane along the line y = 2x, So

Unit vector of length will be = (i + 2j)/|i + 2j|

Unit vector of length = (i + 2j)/sqrt 5

So magnetic force on this wire will be

F2 = 24.0*(0.2225*(i + 2j)/sqrt 5)x(0.318 i)

F2 = (24.0*0.2225*0.318/sqrt 5)*((i + 2j)xi)

F2 = (24.0*0.2225*0.318/sqrt 5)*(ixi + 2*jxi)

F2 = (24.0*0.2225*0.318/sqrt 5)*(0 + 2*(-k))

F2 = (-2*24.0*0.2225*0.318/sqrt 5) k

F2 = -1.519 k

So net force on wire will be

Fnet = F1 + F2

Fnet = -1.698 j - 1.519 k

Magnitude of force will be

|Fnet| = sqrt ((-1.698)^2 + (-1.519)^2)

|Fnet| = 2.28 N

Direction will be given by:

Direction = arctan (-1.519/(-1.698)) = 41.82 deg

theta = 42.0 deg below the -ve y-axis

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