Respuesta :
Answer:
Radius, [tex]r=2.14\times 10^{-5}\ m[/tex]
Explanation:
It is given that,
Magnetic field, B = 0.275 T
Kinetic energy of the electron, [tex]E=4.9\times 10^{-19}\ J[/tex]
Kinetic energy is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2E}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}[/tex]
v = 1037749.04 m/s
The centripetal force is balanced by the magnetic force as :
[tex]qvB\ sin90=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }[/tex]
[tex]r=2.14\times 10^{-5}\ m[/tex]
So, the radius of the circular path is [tex]2.14\times 10^{-5}\ m[/tex]. Hence, this is the required solution.
Answer:
The speed of the electron and the radius of the circular path are [tex]10.38\times10^{5}\ m/s[/tex] and [tex]2.15\times10^{-5}\ m[/tex].
Explanation:
Given that,
Magnetic field = 0.275 T
Kinetic energy [tex]K.E= 4.90\times10^{-19}\ J[/tex]
We need to calculate the speed of the electron
Using kinetic energy
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2K.E}{m}}[/tex]
Where, m = mass of electron
K.E = kinetic energy
Put the value into the formula
[tex]v= \sqrt{\dfrac{2\times4.90\times10^{-19}}{9.1\times10^{-31}}}[/tex]
[tex]v=10.38\times10^{5}\ m/s[/tex]
We need to calculate the radius of the circular path
Using equation of force on charge in magnetic field
[tex]F = qvB\sin\theta[/tex]....(I)
Using centripetal force
[tex]F=\dfrac{mv^2}{r}[/tex]....(II)
From equation (I) and (II)
[tex]\dfrac{mv^2}{r}=qvB\sin\theta[/tex]
[tex]r=\dfrac{mv^2}{qvB\sin\theta}[/tex]
Put the value into the formula
[tex]r=\dfrac{9.1\times10^{-31}\times10.38\times10^{5}}{1.6\times10^{-19}\times0.275\sin90^{\circ}}[/tex]
[tex]r=\dfrac{9.1\times10^{-31}\times10.38\times10^{5}}{1.6\times10^{-19}\times0.275}[/tex]
[tex]r=2.15\times10^{-5}\ m[/tex]
Hence, The speed of the electron and the radius of the circular path are [tex]10.38\times10^{5}\ m/s[/tex] and [tex]2.15\times10^{-5}\ m[/tex].
