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saveliy_v [14]
3 years ago
12

Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is

placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.0 T in 100 ms. Determine the magnitude of the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field.
Physics
1 answer:
Xelga [282]3 years ago
8 0

Answer:

7.3114*10^{-5}V

Explanation:

To give a solution to the exercise, it is necessary to consider the concepts related to magnetic flux and Faraday's law of induction. Faraday's law states that the voltage induced in a closed circuit is directly proportional to the speed with which the magnetic flux that crosses any surface with the circuit as an edge changes over time.

It is represented under the equation,

\epsilon = N \frac{\Delta\Phi}{\Delta t}

Where,

\epsilonis the induced electromotive force

N = Number of loops

\Delta t= Time

\Delta\Phi= Magnetic Flux

For definition the change in magnetic flux is:

\Delta \Phi = \Delta B A cos\phi

Where,

B= Magnetic field

Substituting at the first equation we have

\epsilon = N \frac{\Delta B A Cos\phi}{\Delta t}

\epsilon = N \frac{(B_2-B_1) (\pi r^2) Cos\phi}{\Delta t}

Our values are given by,

N = 1 turn

B_2 = 1T

B_1 = 0T

r = 1.6mm

\phi = 0\°

\Delta t = 100ms

Replacing,

\epsilon = (1) \frac{(1-0) (\pi (1.6*10^{-3})^2) Cos(0)}{110*10^{-3}}

\epsilon = 7.311*10^{-5}V

<em>Therefore the magnitud of the induced emf around a horizontal circle of tissue is 7.3114*10^{-5}V</em>

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A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . A
yaroslaw [1]

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

v_{f}^{2}=v_{0}^{2}+2gy\\v_{f}=\sqrt{0+2(9.8\frac{m}{s^{2}})(145m)}=53.31\frac{m}{s}

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

F=(3.8*10^{8})(1.602*10^{-19}C)(53.31\frac{m}{s})(0.205T)=6.65*10^{-10}N

B)

b.  south - north (by the rigth hand rule)

I hope this is usefull for you

regards

8 0
3 years ago
The 8.00-cm long second hand on a watch rotates smoothly.
dolphi86 [110]
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4 0
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How permeable and porous would an aquifer be?
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Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l
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The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
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The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

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= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

8 0
4 years ago
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