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ad-work [718]
3 years ago
13

8) Now drop the magnetic ball of the same size down the section of copper tube. Currents are induced in the tube both above and

below the falling magnet (pictured here as a disc for clarity). On the drawing below, find the direction of the induced current, the direction of the induced magnetic field, and the N/S poles for both induced currents. Explain how these induced currents slow down the falling magnet.
Physics
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

Explanation:

To have a deeper understanding of this. Let's begin with the acceleration due to gravity (g), with N pointing downward.

Since magnetic field lines originate in the north, field strength decreases for the upper disk and increases for the lower disk when the disk is in the center of two disks.

According to Lenz's theorem, current-induced would be so that the magnetic field due to this induced current opposes the magnetic field due to which there is an induced current, so induced current will be clockwise for the upper disk and anticlockwise for the lower disk.

If the south pole is pointing backward, the generated current will flow anticlockwise for the upper disk and clockwise for the lower disk.

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Answer:

a) 1.2*10^{-3}cos(1.25t)

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Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

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