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matrenka [14]
3 years ago
12

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 90 mT, and the electrons' kinetic energy is 2.0 eV.If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Physics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

Explanation:

Electron's kinetic energy = 2 eV

= 2 x 1.6 x 10⁻¹⁹ J

1/2 m v² = 3.2 x 10⁻¹⁹

1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹

v² = .703 x 10¹²

v = .8385 x 10⁶ m/s

Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows

r = mv / Bq

where m , v and q are mass , velocity and charge of electron .

here given magnetic field B = 90 mT

= 90 x 10⁻³ T

Putting these values in the expression above

r =  mv / Bq

= \frac{9\times10^{-31}\times.8385\times10^6}{90\times10^{-3}\times1.6\times10^{-19}}

= .052 mm.

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Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

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E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

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E⃗ 2

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Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

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