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Ira Lisetskai [31]
3 years ago
14

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil

makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A
Physics
1 answer:
Travka [436]3 years ago
8 0

Complete question:

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A.

What is magnitude of the magnetic moment of the coil? Answer in units of A · m2.

Answer:

The magnetic moment of the coil is 0.0252 A.m²

Explanation:

Given;

radius of the coil, r = 4 cm = 0.04 m

number of turns of the coil, N = 5 turns

magnetic field strength B = 0.8 T

current in the coil, I = 1 A

Area of the coil, A = πr² = π(0.04)² = 0.00503 m²

magnetic moment of the coil, μ = NIA

where;

N is the number of turns

I is the current in the coil

A is the area of the coil

magnetic moment of the coil, μ = 5 x 1 x 0.00503 = 0.0252 A.m²

Therefore, the magnetic moment of the coil is 0.0252 A.m²

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Answer:

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Explanation:

Parameters given:

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We know that Electrostatic force, F, is given in terms of Electric field, E, as:

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The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

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Equating the two forces of the object, we get:

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Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

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