physics

a current-carrying wire is oriented at right angles to a uniform magnetic field. if the lenght of the wire is 0.50m and it experiences a force at 0.90n while carrying a current of 4.0a. what is the strength of the magnetic field?

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asked by mj
  1. The answer is .45 T
    F = B(I)L
    B = F/(I)L
    B = .90/4(.5)
    B = .45 T
    (T is Teslas)

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    posted by TJ

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