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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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Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
When lights are wired such that if one bulb fails, the entire string of lights
Viktor [21]

Answer:

series

Explanation:

In a series circuit all the components are attached to one branch, so that if one component fails, all the others stop working. In a parallel circuit, however, the components are wired in separate branches, so that even if one branch fails, the rest are not disrupted.

8 0
2 years ago
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A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect fricti
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Answer:

1.66 m/s

Explanation:

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4864=\frac{1}{2} (3540)v^{2}

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6 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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3 0
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What is the difference between a free body diagram and a vector diagram?​
Murrr4er [49]

Answer:

Free body diagrams are used to describe situations where several forces act on an object. On the other hand Vector diagrams are used to resolve (break down) a single force into two forces acting as right angles to eachother

Explanation:

Hope this helps !

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