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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? 0.0005 T 0.0085 T 0.0013 T 0.0001 T What is the direction of the magnetic field? +x direction −x direction +y direction −y direction +z direction −z direction

Respuesta :

Explanation:

It is given that,

Current carried in the conductor, I = 14.7 A (+x axis)

The magnetic force per unit length on the conductor, [tex]\dfrac{F}{L}=0.125\ N/m[/tex] (-y axis)

The magnetic force is given by :

[tex]F=ILBsin\theta[/tex]

[tex]\theta=90[/tex]

[tex]F=ILB[/tex]

[tex]B=\dfrac{F}{IL}[/tex]

[tex]B=\dfrac{0.125}{14.7}[/tex]  

B = 0.0085 T

According to right hand rule, the direction of magnetic force can be calculated. So, the direction of magnetic field is along +z axis. Hence, this is the required solution.

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