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Tatiana [17]
3 years ago
6

A loop of wire is placed in a uniform magnetic field. (a) For what orientation of the loop is the magnetic flux a maximum? The p

lane of the loop is perpendicular to the magnetic field. The plane of the loop is parallel to the magnetic field. The magnetic flux is independent of the orientation of the loop. (b) For what orientation is the flux zero? The plane of the loop is perpendicular to the magnetic field. The plane of the loop is parallel to the magnetic field. The magnetic flux is independent of the orientation of the loop.
Physics
1 answer:
ra1l [238]3 years ago
5 0

Answer:

1) The plane of the loop is perpendicular to the magnetic field.

2) The magnetic flux is independent of the orientation of the loop.p

Explanation:

The flux is calculated as φ=BAcosθ. The flux is therefore the highest when the magnetic field vector is perpendicular to the plane of the loop We can also deduce that the flux is zero when there is no magnetic field part perpendicular to the loop When the angle reaches zero, the flux is in the limit because when the angle becomes zero, the cos is the maximum.

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A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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