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A wire carries a steady current of 2.80 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, = 1.50 T. If the current is in the positive x direction, what is the magnetic force on the section of wire?

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samuelonum1

Answer:

The magnetic force in the wire is 3.15N  

Explanation:

Given

current I= 2.80 A.

length of conductor L= 0.75 m

Magnetic field, B = 1.50 T

∅=90

according to Fleming's left hand rule the conductor will observe a force perpendicular to it

Applying the formula

[tex]F= BIL* sin(90)[/tex]

[tex]F=1.50* 2.80* 0.750* sin(90)\\\F= 3.15N[/tex]

Note: sine(90)= 1

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