Question

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

Answer 1

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 $\vec {F} = i (\vec {l} x \vec {B})$