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A current loop, carrying a current of 18.0 A, is in the shape of a right triangle with sides 30, 40, and 50 cm. The loop is in a uniform magnetic field of magnitude 100 mT whose direction is parallel to the current in the 50 cm side of the loop. Find the magnitude of each of the following.(a) the magnetic dipole moment of the loop A·m2(b) the torque on the loop N·m

Answer :

Answer:

The magnetic moment is 1.08 [tex]A.m^{2}[/tex] and torque on the loop is [tex]108 \times 10^{-3}[/tex] N.m

Explanation:

Given:

Magnetic field [tex]B = 100 \times 10^{-3}[/tex] T

Current [tex]I = 18[/tex] A

Area of right triangle [tex]A = \frac{1}{2} \times (40 \times 30) \times 10^{-4 } = 0.06 m^{2}[/tex]

(A)

From the formula of magnetic dipole,

   [tex]\mu = IA[/tex]

   [tex]\mu = 18 \times 0.06[/tex]

   [tex]\mu = 1.08[/tex] [tex]Am^{2}[/tex]

(B)

The torque on the loop is,

   Γ = [tex]\mu B[/tex]

   Γ = [tex]1.08 \times 100 \times 10^{-3}[/tex]

   Γ = [tex]108 \times 10^{-3 }[/tex] N.m

Therefore, the magnetic moment is 1.08 [tex]A.m^{2}[/tex] and torque on the loop is [tex]108 \times 10^{-3}[/tex] N.m

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