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A current of 15.0 mA is maintained in a single circular loop with a circumference of...

A current of 15.0 mA is maintained in a single circular loop with a circumference of 3.00 m. A magnetic field of 0.750 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?
N · m

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Answer #1

Given that current i=15*10^-3 A

circumfrence of the loop is S=3.0m

hence the radius is r=S/2pi

                               =0.477m

    magnetic field strength is B=0.750T

a)magnetic moment of the loop is

M=iA

    =i(pir^2)

=15*10^-3(3.14*0.477^2)

    =1.07 mA m2

b)torque T=iAB

                  =i(pi r^2 )B

=15*10^-3(3.14*0.477^2)0.750

                   =0.8032 mN m

I hope help you !!

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