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stepladder [879]
3 years ago
14

An equilateral triangle 8.0 cm on a side is in a 3.5 mT uniform magnetic field. The magnetic flux through the triangle is 6.0 μW

b. What is the angle between the magnetic field and an axis perpendicular to the plane of the triangle?
Physics
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

43.80°

Explanation:

The side of the equilateral triangle is given as 8 cm

The area of the equilateral triangle A=\frac{\sqrt{3}}{4}a^2=\frac{1.732\times 8^2}{4}=27.712cm^2=27.712\times 10^{-4}m^2

Magnetic field B=3.5mT=3.5\times 10^{-3}T

Magnetic flux \Phi =6\mu Wb=6\times 10^{-6}weber

We know that  \Phi =BACOS\Theta

So 6\times 10^{-6}=3\times 10^{-3}\times 27.712\times 10^{-4}cos\Theta

cos\Theta =0.7217

\Theta =43.80^{\circ}

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The velocity as a function of time t is:

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The problem asks as the time t at which U=3K, so we have:

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So, the time needed is 3 seconds.

d) 0.097 m

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T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

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L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






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