A 5.0-cm length of wire carries a current of 2.0 A in the negative x-direction. The force on the wire when it is in a magnetic field B is 1.5j – 2.5k. When the wire is rotated 90o in the x-y plane so that the current is in the positive y-direction, the force on it is -1.2k. Find the magnetic field B.
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OK I'm only doing this question cuz I've got an exam on this stuff so I'm practicing. :P
(current in wire) = I
(length of wire) = L = 0.05 m
(external magnetic field) = B = (B_x) [i] + (B_y) [j] + (B_z) [k]
(magnetic force on wire) = F
F = L·(I x B)
B = (B_x) [i] + (B_y) [j] + (B_z) [k]
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PART ONE:
I = (- 2.0 A) [i]
F = (1.5 N) [j] + (- 2.5 N)[k]
Based on the given information, we can find the y- and z-components of the magnetic field but not the x-component yet. The wire is currently parallel to the x-component of B.
F = L·(I x B)
Do the cross product (I x B), with I = (- 2.0 A) [i] and B = (B_x) [i] + (B_y) [j] + (B_z)[k]
Then you'll find that (I x B) = (2.0 A)(B_z)[j] + (- 2.0 A)·(B_y)[k]
F = L·(I x B)
(1.5 N) [j] + (- 2.5 N)[k] = (0.10 A·m)(B_z)[j] + (- 0.10 A·m)·(B_y)[k]
Therefore,
B_z = (1.5 N)/(0.10 A·m) = 15.0 T
B_y = (- 2.5 N)/(- 0.10 A·m) = 25.0 T
B_x is still unknown... it is found in Part 2.
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PART TWO:
I = (2.0 A) [j]
F = (-1.2 N) [k]
The current is now in the positive y-direction - which importantly, means that it's no longer parallel to the x-direction. Thus, we'll find B_x.
(I x B) = (2.0 A)(B_z) [i] - (2.0 A)(B_x) [k]
F = L·(I x B)
(-1.2 N) [k] = (0.10 A·m)(B_z) [i] + (- 0.10 A·m)(B_x) [k]
Thus,
B_x = (-1.2 N)/(-0.10 A·m) = 12.0 T
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Therefore, B = (12.0 T) [i] + (25.0 T) [j] + (15.0 T) [k]
(current in wire) = I
(length of wire) = L = 0.05 m
(external magnetic field) = B = (B_x) [i] + (B_y) [j] + (B_z) [k]
(magnetic force on wire) = F
F = L·(I x B)
B = (B_x) [i] + (B_y) [j] + (B_z) [k]
-----
PART ONE:
I = (- 2.0 A) [i]
F = (1.5 N) [j] + (- 2.5 N)[k]
Based on the given information, we can find the y- and z-components of the magnetic field but not the x-component yet. The wire is currently parallel to the x-component of B.
F = L·(I x B)
Do the cross product (I x B), with I = (- 2.0 A) [i] and B = (B_x) [i] + (B_y) [j] + (B_z)[k]
Then you'll find that (I x B) = (2.0 A)(B_z)[j] + (- 2.0 A)·(B_y)[k]
F = L·(I x B)
(1.5 N) [j] + (- 2.5 N)[k] = (0.10 A·m)(B_z)[j] + (- 0.10 A·m)·(B_y)[k]
Therefore,
B_z = (1.5 N)/(0.10 A·m) = 15.0 T
B_y = (- 2.5 N)/(- 0.10 A·m) = 25.0 T
B_x is still unknown... it is found in Part 2.
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PART TWO:
I = (2.0 A) [j]
F = (-1.2 N) [k]
The current is now in the positive y-direction - which importantly, means that it's no longer parallel to the x-direction. Thus, we'll find B_x.
(I x B) = (2.0 A)(B_z) [i] - (2.0 A)(B_x) [k]
F = L·(I x B)
(-1.2 N) [k] = (0.10 A·m)(B_z) [i] + (- 0.10 A·m)(B_x) [k]
Thus,
B_x = (-1.2 N)/(-0.10 A·m) = 12.0 T
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Therefore, B = (12.0 T) [i] + (25.0 T) [j] + (15.0 T) [k]
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You're given all the information you need; since the currents are steady, you can use
F = IL x B
Just make sure you get the components right ;)
F = IL x B
Just make sure you get the components right ;)
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Physics II: I never took it.