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Julli [10]
3 years ago
13

A conductor carrying 14.7 amps of current is directed along the positive x-axis and perpendicular to a uniform magnetic field. A

magnetic force per unit length of 0.125 N/m acts on the conductor in the negative y direction. What is the strength of the magnetic field at the place where the current is? 0.0013 T 0.0001 T 0.0085 T 0.0005 T What is the direction of the magnetic field? +x direction −x direction +y direction −y direction +z direction −z direction
Physics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

B = 0.00782N / (Am) = 7.82 * 10^{-3} T

Explanation:

Under the definition of force on a magnetic field we proceed to make the point a,

A) F = ILBsin (\theta)

Since the field is perpendicular, then \theta = 0

F = ILB

B = \frac {F} {IL} = \frac {1} {I} \frac {f} {L}

B = 0.00782N / (Am) = 7.82 * 10^{-3} T

b) The direction of the magnetic field would be done in + Zdirection. This when understanding that the vector extends along the z + axis of the formula \vec {F} = i (\vec {l} x \vec {B})

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