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A wire carries a current of 6.52 A in a direction that makes an angle of 44.8 degrees with the direction of a magnetic field of strength 0.258 T. What is the magnetic force on a 3.61 m length of wire

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okpalawalter8

Answer:

[tex]F=4.27N[/tex]

Explanation:

From the question we are told that:

Current  [tex]I = 6.52 A[/tex]

Angle [tex]\theta= 44.8 \textdegree[/tex]

Magnetic field   [tex]B= 0.258 T.[/tex]

Length [tex]l=3.61m[/tex]

Generally the equation for  Magnetic force F is mathematically given by

 [tex]F=BIlsin\theta[/tex]

Therefore

 [tex]F=0.258*6.52*3.61*sin 44.8[/tex]

 [tex]F=4.27N[/tex]

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