Respuesta :
Answer:
[tex]B_e = 2.72 \times 10^{-4} T[/tex]
Explanation:
As we know that when charge particle is projected in perpendicular magnetic field then the radius of the charge particle is given as
[tex]F = qvB[/tex]
[tex]\frac{mv^2}{r} = qvB[/tex]
now we have
[tex]r = \frac{mv}{qB}[/tex]
since here radius of proton and electron will be same
so we will have
[tex]r_e = r_p[/tex]
[tex]\frac{m_e v}{q_e B_e} = \frac{m_p v}{q_p B_p}[/tex]
so we have
[tex]B_e = \frac{B_p m_e}{m_p}[/tex]
given that
[tex]B_p = 0.50 T[/tex]
[tex]m_e = 9.11 \times 10^{-31} kg[/tex]
[tex]m_p = 1.67 \times 10^{-27} kg[/tex]
so we have
[tex]B_e = \frac{0.50(9.11\times 10^{-31})}{1.67\times 10^{-27}}[/tex]
[tex]B_e = 2.72 \times 10^{-4} T[/tex]
Answer:
2.73 x 10^-4 T
Explanation:
The relation between the radius of circular path and the velocity is given by
[tex]r=\frac{mv}{Bq}[/tex]
Where, m be the mass of charged particle and q be the charge on the charged particle.
For proton:
mass of proton = mp = 1.67 x 10^-27 kg
charge of proton = q = 1.6 x 10^-19 C
magnetic field strength, B = 0.50 T
[tex]r=\frac{1.67\times 10^{-27}v}{0.5 \times 1.6 \times 10^{-19}}[/tex] ..... (1)
For electron:
mass of electron = me = 9.1 x 10^-31 kg
charge of electron = q = 1.6 x 10^-19 C
let the strength of magnetic field is B.
[tex]r=\frac{9.1\times 10^{-31}v}{B \times 1.6 \times 10^{-19}}[/tex] ..... (2)
As the radius and the velocity is same for both the particles, so by comparing equation (1) and equation (2), we get
[tex]\frac{1.67\times 10^{-27}v}{0.5 \times 1.6 \times 10^{-19}} = \frac{9.1\times 10^{-31}v}{B \times 1.6 \times 10^{-19}}[/tex]
[tex]B = \frac{9.1\times10^{-31}\times0.5}{1.67\times10^{-27}}[/tex]
B = 2.73 x 10^-4 T
Thus, the magnetic field required for electron is 2.73 x 10^-4 T.
