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Blababa [14]
2 years 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:
GrogVix [38]2 years ago
5 0

Explanation:

It is given that, the field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs

Change in magnetic field, dB=2.5\ T

Change in time, dt=200\ \mu s=200\times 10^{-6}\ s=2\times 10^{-4}\ s

Diameter, d = 2.3 cm

Radius, r = 0.0115 m

Emf is induced in the ring as the field changes. It is given by :

E=\dfrac{d\phi}{dt}

E=\dfrac{d(B.A\ cos(0))}{dt}

E=A\dfrac{d(B)}{dt}

E=\pi (0.0115)^2\dfrac{2.5}{2\times 10^{-4}}

E = 5.19 volts

So, the emf induced in the ring is 5.19 volts. Hence, this is the required solution.

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Neutron: v=0.688 m/s

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h is the Planck constant: 6.626×10⁻³⁴\frac{kg.m^2}{s}

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m=1.673×10⁻²⁴ g · \frac{1kg}{1000g}=1.673×10⁻²⁷ kg

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v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)

v=0.689 m/s

<u>Neutron:</u>

m=1.675×10⁻²⁴ g · \frac{1kg}{1000g}=1.675×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)

v=0.688 m/s

<u>Electron:</u>

m= 9.109×10⁻²⁸ g · \frac{1kg}{1000g}=9.109×10⁻³¹ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(9.109×10⁻³¹ kg×575×10⁻⁹m)

v=1265.078 m/s

<u>Alpha particle:</u>

m=6.645×10⁻²⁴ g · \frac{1kg}{1000g}=6.645×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)

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