Respuesta :
Answer:
The magnetic field strength is 0.086 T
The maximum kinetic energy is [tex]1.62 \times 10^{-14}[/tex] J
Explanation:
Given:
Frequency [tex]f = 2.4 \times 10^{9}[/tex] Hz
(A)
The magnetic field related to cyclotron is given by,
[tex]B = \frac{2\pi mf}{e}[/tex]
Where [tex]m =[/tex] mass of electron [tex]= 9.1 \times 10^{-31 }[/tex] kg, [tex]e = 1.6 \times 10^{-19}[/tex] C
[tex]B = \frac{6.28 \times 9.1 \times 10^{-31 }\times 2.4 \times 10^{9} }{1.6 \times 10^{-19} }[/tex]
[tex]B = 0.086[/tex] T
Therefore, the magnetic field strength is 0.086 T
(B)
Diameter of orbit [tex]d= 2.5 \times 10^{-2}[/tex] m
Radius of orbit [tex]r = 1.25 \times 10^{-2}[/tex] m
The maximum kinetic energy is given by,
KE = [tex]\frac{1}{2} mv^{2}[/tex]
Where [tex]v = r \omega = r 2\pi f[/tex]
KE = [tex]\frac{1}{2} m(2\pi fr)^{2}[/tex]
KE = [tex]\frac{1}{2} \times 9.1 \times 10^{-31 } \times (2 \times 3.14 \times 2.4 \times 10^{9} \times 1.25 \times 10^{-2} )^{2}[/tex]
KE = [tex]1.62 \times 10^{-14}[/tex] J
Therefore, the maximum kinetic energy is [tex]1.62 \times 10^{-14}[/tex] J
A) The magnetic field strength of the magnetic field is; B = 0.0856 T
B) The maximum electron kinetic energy is; KE = 1.617 × 10^(-14) J
A) The formula for Magnetic field as it relates to cyclotron is;
B = 2πmf/e
Where;
B is magnetic field
m is mass of electron = 9.1 × 10^(-31) kg
F is frequency
e is charge on electron = 1.6 × 10^(-19) C
Since we are given f = 2.4 GHz = 2.4 × 10^(9) Hz, then;
B = (2π × 9.1 × 10^(-31) × 2.4 × 10^(9))/(1.6 × 10^(-19))
B = 0.0856 T
B) We are told that the maximum diameter of the orbit is 2.5 cm. Thus;
d = 2.5 cm = 0.025 m
radius; r = d/2 = 0.025/2
r = 0.0125 m
Now, formula for kinetic energy is;
KE = ½mv²
Where v = 2πrf
KE = ½m(2πrf)²
KE = 2mπrf
KE = 2 × 9.1 × 10^(-31) × π × 0.0125 × 2.4 × 10^(9)
KE = 1.617 × 10^(-14) J
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