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777dan777 [17]
3 years ago
7

A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is m

ounted on an axis, perpendicular to the magnetic field, which allows the loop to rotate. What is the torque on the loop when its plane is oriented at a 21° angle to the field?
Physics
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:

The torque on the loop is 2.4 \times 10^{-2} Nm

Explanation:

Given:

Current I = 1.6 A

Magnetic field B = 0.30 T

Area of loop A = 0.14 m^{2}

Angle between magnetic field and area vector \theta = 21°

Form the formula of  torque in case of magnetic field,

 г = MB \sin \theta

Where M = magnetic moment

  M = IA

 г = IAB \sin 21

 г = 1.6 \times 0.30 \times 0.14 \times 0.3583

 г =2.4 \times 10^{-2} Nm

Therefore, the torque on the loop is 2.4 \times 10^{-2} Nm

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Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

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