would you like answer those 3 question please 1. A wire 2.0 m long is suspended
ID: 1560900 • Letter: W
Question
would you like answer those 3 question please
1. A wire 2.0 m long is suspended parallel to a uniform magnetic field of 0.50 T. If the current in the wire is 0.60 A, what is the magnetic force in N on the wire? answer 0
2. A particle with a charge 3.2 X 10-19 C is moving at a speed of 5x107 m/s that makes an angle of 40o with respect to a uniform magnetic field B. The particle experiences a magnetic force of magnitude 2.47 X 10-14 N. Determine the magnetic field strength B. Answer: 2.4 mT
3. If a current-carrying conductor has no magnetic force on it when placed in a constant magnetic field,
the wire is __________ to the field. Answer: parallel
the wire is __________ to the field. Answer: parallel
Explanation / Answer
1)
F = B*I*L*sin(theta) (here theta is the angle between I and B)
= B*I*L*sin(0)
= 0
2)
use, F = q*v*B*sin(theta)
==> B = F/(q*v*sin(theta))
= 2.47*10^-14/(3.2*10^-19*5*10^7*sin(40))
= 2.4*10^-3 T
= 2.4 mT
3) F = B*I*L*sin(theta)
0 = B*I*L*sin(theta)
==> sin(theta) = 0
==> theta = 0
so, the wire must be parellel to the manetic field.
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