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grandymaker
15 days ago
12

A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-

field machine rises from 0 T to 2.5 T in 200 μs. Suppose a technician holds his hand near the device so that the axis of his 2.3-cm-diameter wedding band is parallel to the field. Part A What emf is induced in the ring as the field changes
Physics
1 answer:
Sav [3.1K]15 days ago
5 0
It is stated that, in a typical pulsed-field machine, the magnetic field rises from 0 T to 2.5 T within 200 μs. The change in the magnetic field and time interval are relevant here. The diameter is 2.3 cm, translating to a radius of 0.0115 m. As the magnetic field changes, an induced emf occurs within the ring, determined by: E = 5.19 volts. Thus, the induced emf in the ring equates to 5.19 volts, which is the sought solution.
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A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold
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Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Insert the values into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We will compute the coefficient of permeability

Applying the formula for permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability measures as 8.6\times10^{-3}\ cm/s.

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

v=ki

v=\dfrac{kh}{l}

Substituting the values into the equation

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

The discharge velocity during the test measures 0.0187 cm/s.

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(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

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