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A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50×10−5T and points into the Earth at an angle of 56.0∘ below a line pointing due north. A 7.80-A clockwise current passes through the coil.

Determine the torque on the coil.

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onyebuchinnaji

Answer:

The torque in the coil is  4.9 × 10⁻⁔ N.m  

Explanation:

T = NIABsinΞ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁔ T

Ξ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2[/tex]

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁔ × sin34°

T = 4.9 × 10⁻⁔ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁔ N.m

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