Let's apply the torque equation to a circular coil of wire in a magnetic field. The coil has an average radius of 0.0435 m , has 40 turns, and lies in a horizontal plane. It carries a current of 7.25 A in a counterclockwise sense when viewed from above. The coil is in a uniform magnetic field directed toward the right, with magnitude 1.05 T . Find the magnetic moment and the torque on the coil. Which way does the coil tend to rotate

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okpalawalter8 okpalawalter8 · Answer 1 · 2021-05-15T03:00:42+02:00

Answer:

[tex]M=1.72Am^2[/tex]

[tex]T=1.810N-m[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.0435[/tex]

Number of turns [tex]N=40turns[/tex]

Current [tex]I= 7.25 A[/tex]

Magnetic Field [tex]B=1.05T[/tex]

Generally the equation for magnetic moment M is mathematically given by

[tex]M=NIA[/tex]

Where

[tex]A= Surface Area[/tex]

[tex]A=\pi(r)^2[/tex]

[tex]A=3.142*(0.0435)^2[/tex]

[tex]A=5.95*10^{-3}[/tex]

Therefore the magnetic moment is given as

[tex]M=NIA[/tex]

[tex]M=40*7.25*5.95*10^{-3}[/tex]

[tex]M=1.72Am^2[/tex]

Generally the equation for Torque on the coil T is mathematically given by

[tex]T=M*B[/tex]

[tex]T=1.72*1.05[/tex]

[tex]T=1.810N-m[/tex]

Therefore torque on coil T is given a

[tex]T=1.810N-m[/tex]