A circular wire loop with a radius of 66.5~cm66.5 cm lies flat in the horizontal x-y plane and contains a current of 847~mA847 mA that runs counter-clockwise in the wire loop when viewed from above. If the loop sits in a uniform magnetic field of 3.5~T3.5 T that is oriented in the horizontal x direction, what is the magnitude of the torque on the wire loop

the magnitude of the torque on the wire loop is 4.1185 N.m

Explanation:

Given the data in the question;

Area of circular loop A = πr²

given that radius = 66.5 cm = 0.665 m

Area of circular loop A = π( 0.665 m )² = 1.38929 m²

Magnitude of Torque T = M × B × sin∅

where M is the magnetic moment

∅ is the angle between M¬ and B¬

given that mA847 mA = 0.847 A

so |M¬| = 0.847 × 1.38929 m² = 1.1767; directed  vertically upward

given that the loop sits in a uniform magnetic field of 3.5 T that is oriented in the horizontal x direction

B¬ = 3.5 T (i^)

so ∅ between M¬ and B¬ is 90°

from the initial formula

Magnitude of Torque T = M × B × sin∅

we substitute in our values;

Magnitude of Torque T = 1.1767 × 3.5 × sin90°

= 1.1767 × 3.5 × 1

= 4.1185 N.m

Therefore, the magnitude of the torque on the wire loop is 4.1185 N.m

the answer is a)

mark brainliest

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