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3 years ago

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A simple electric motor consists of a 220-turn coil, 4.4 cm in diameter, mounted between the poles of a magnet that produces a 9

6-mT field. Part A When a 15-A current flows in the coil, what is the coil's magnetic dipole moment?

1 answer:

Verdich [7]3 years ago

4 0

Answer:

$M = 5.01\ A.m^2$

Explanation:

It is given that,

Number of turns in the coil, N = 220

Diameter of the coil, d = 4.4 cm

Radius of the coil, r = 2.2 cm = 0.022 m

Magnetic field produced by the poles of magnet, $B=96\ mT=96\times 10^{-3}\ T$

Current flowing in the coil, I = 15 A

Let M is the coil's magnetic dipole moment. Its formula is given by :

$M=N\times I\times A$

$M=220\times 15\times \pi (0.022)^2$

$M = 5.01\ A.m^2$

So, the coil's magnetic dipole moment is $5.01\ A.m^2$ . Hence, this is the required solution.

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

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a) $x(t)=2.0 sin (10 t) [m]$

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$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

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$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

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:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

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with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

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c) 3 seconds.

When x=0, the equation of motion is:

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so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

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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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