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Tema [17]
2 years ago
14

A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are

oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being perpendicular to the magnetic field.
A)What is the magnitude of the magnetic moment of the loop?
B)What torque does the magnetic field exert on the loop?
Physics
1 answer:
Dahasolnce [82]2 years ago
4 0

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

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Here we have:

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Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

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Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

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Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

M=\frac{h_i}{h_o}=-\frac{q}{p}

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p = 5.0 m is the distance of the object (the man) from the lens

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A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10
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I= \frac{1}{2} c \epsilon_0 E^2
where
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By substituting the numbers of the problem and re-arranging the equation, we can find E:
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F=qvB \sin \theta
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2 years ago
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