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A flat rectangular loop of wire carrying a 4.0-a current is placed in a uniform 0.60-t magnetic field. the magnitude of the torque acting on this loop when the plane of the loop makes a 30° angle with the field is measured to be 1.1 n ∙ m. what is the area of this loop?

Respuesta :

Torque on a closed current carrying loop is given by formula

[tex]T = NiABsin\theta[/tex]

here

N = number of turns = 1

i = current = 4 A

A = area of the loop

B = magnetic field = 0.60 T

[tex]\theta[/tex] = angle between normal of the plane and magnetic field = 90 - 30 = 60 degree

Now we will have

[tex]1.1 = 1* 4 * A * 0.60* sin60[/tex]

[tex] 1.1 = 2.078*A[/tex]

[tex]A = 0.53 m^2[/tex]

So area of the loop is 0.53 m^2


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