Types of magnetic resonance image

An arthroscope can be inserted into a knee joint through a very tiny incision to help a doctor examine the knee. A light is tran

natali 33 [55]

Sample Response: The technology being described is fiber optics. Since they are small and flexible, they enable doctors to see areas that they might otherwise be unable to see without surgery.

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3 years ago

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Although the evidence is weak, there has been concern in recent years over possible health effects from the magnetic fields gene

Natasha2012 [34]

Answer:

A. $B = 6.36 * 10^{-10} T$

B. P ≈ 0

Explanation:

In order to calculate the magnetic field strength we have to use the magnetic field strength of a straight wire.

$B = \frac{mi* I}{2\pi *d}$ (eq. I)

B = magnetic field strength at distance d

I = current (A)

mi = represented by the greek letter μ, represents the permeability of the free space, which is: 4 × π 10^(-7) T m/A

d = distance from the wire

By replacing the values in eq I, we have the following:

$B = \frac{4\pi 10^{-7} T m A^{-1} 200 A}{2\pi *20 m}\\\\B = 6.36 * 10^{-10} T\\$ (eq II)

The earth magnetic field in the surface variates from 25 to 65 microteslas. Thus:

P = Percentage from the wires/percentage of the earth

$P = \frac{6.36 * 10^{-10}T}{65* 10^{-3} T}\\$ ∵ $B$ ∴

P ≈ 0

5 0

4 years ago

Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s

vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

$\Delta y = \frac{x\lambda}{d}$

Where

$\Delta y$ = Distance between fringes

x = distance between slits and screen

d = Distance between slits

$\lambda$ = Wavelength

Our values are given as

d= 0.08mm

x =3m

$\lambda = 633nm$

In this way replacing in the equation,

$\Delta y = \frac{x\lambda}{d}$

$\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}$

$\Delta y = 2.37*10^{-2}m$

$\Delta y = 2.37cm$

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

$\Delta Y_2 = \frac{\Delta Y_1}{n}$