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Calculate the radius of the circular orbit for the parameters specified below. Part (a) uses parameter values that can be set using the animation sliders, and you can use the animation to verify your calculations. Part (b) uses parameter values outside the range of the animation. (a) Let the magnitude of the magnetic field B = 0.9 T and let the charge be an electron moving with a speed of 2.2 ✕ 10^6 m/s. r = (b) Let the magnitude of the magnetic field B = 1.2 T and let the charge be an electron moving with a speed of 7.3 ✕ 10^6 m/s. r =

Respuesta :

Answer:

a)[tex]r=1.3933\times 10^{-5}\,m[/tex]

b)[tex]r=3.4675\times 10^{-5}\,m[/tex]

Explanation:

(a)

Given that:

magnetic field,[tex]B=0.9\,T[/tex]

velocity of an electron,[tex]v=2.2\times 10^{6}\,m.s^{-1}[/tex]

we know the charge on an electron,

[tex]q=1.6\times 10^{-19}\,C[/tex]

Since the direction of velocity and magnetic field are not mentioned therefore we assume it to be mutually perpendicular to each other so that it makes a circular trajectory.

For such a case, we have the formula of radius as:

[tex]r=\frac{m.v}{q.B}[/tex]

where m= mass of the charge.

here [tex]m=9.12\times 10^{-31}\,kg[/tex]

[tex]r=\frac{9.12\times 10^{-31}\times 2.2\times 10^{6}}{1.6\times 10^{-19}\times 0.9}[/tex]

[tex]r=1.3933\times 10^{-5}\,m[/tex]

(b)

magnetic field,[tex]B=1.2\,T[/tex]

velocity of an electron,[tex]v=7.3\times 10^{6}\,m.s^{-1}[/tex]

we know the charge on an electron,

[tex]q=1.6\times 10^{-19}\,C[/tex]

Since the direction of velocity and magnetic field are not mentioned therefore we assume it to be mutually perpendicular to each other so that it makes a circular trajectory.

For such a case, we have the formula of radius as:

[tex]r=\frac{m.v}{q.B}[/tex]

where m= mass of the charge.

here [tex]m=9.12\times 10^{-31}\,kg[/tex]

[tex]r=\frac{9.12\times 10^{-31}\times 7.3\times 10^{6}}{1.6\times 10^{-19}\times 1.2}[/tex]

[tex]r=3.4675\times 10^{-5}\,m[/tex]