. A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and
1 answer:
Answer:
The resulting magnetic force on the wire is -1.2kN
Explanation:
The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is
F = I (L*B)
Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)
L x B = = (0, 0, -80)
we can now solve
F = I (L x B) = I (-80)
F = -1200 kmN
F = -1200 kN * 10⁻³
F = -1.2kN
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Answer:
Approximately , assuming friction between the vehicle and the ground is negligible.
Explanation:
Let denote the mass of the vehicle. Let
denote the initial velocity of the vehicle. Let
denote the spring constant (needs to be found.) Let
denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
.
Initial kinetic energy () of the vehicle:
.
When the vehicle is brought to a rest, the elastic potential energy () stored in the spring would be:
.
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial of the vehicle should be equal to the
of the vehicle. In other words:
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Rearrange this equation to find an expression for , the spring constant:
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Substitute in the given values ,
, and
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