A crossed-field velocity selector has a magnetic field of magnitude 0.03 T. The mass of the electron is 9.10939 × 10−31 kg. What electric field strength is required if 6.5 keV electrons are to pass through undeflected?

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mavila18 mavila18 · Answer 1 · 2020-04-08T04:44:27+02:00

Answer:

E = 4.78*10^6 N/C

Explanation:

In this case you have to take into account that both magnetic force and electric force must be equal

[tex]F_E=F_B\\\\qE=qvB[/tex]

[tex]E=vB[/tex]

However, you have to know first what is the speed of the electron by using the expression for the kinetic energy (1keV=1000*1.6*10^{-19}J):

[tex]E_k=\frac{1}{2}m_ev^2\\\\v=\sqrt{\frac{2E_k}{m_e}}=\sqrt{\frac{2(6500(1.6*10^{-19}J))}{9.1*10^{-31}kg}}=4.78*10^{7}\frac{m}{s}[/tex]

by replacing in the expression for the forces you have

[tex]E=(4.78*10^{7}\frac{m}{s})(0.03T)=1.43*10^{6}N/C[/tex]

HOPE THIS HELPS!!

Cricetus Cricetus · Answer 2 · 2022-04-05T17:04:26+02:00

The required electric field strength will be "1.43 × 10⁶ N/C". To understand the calculation, check below.

According to the question,

Mass of electron, [tex]m_e[/tex] = 9.1 × 10⁻³¹ kg

Magnetic field, B = 0.03 T

We know the relation,

→ [tex]E_k[/tex] = [tex]\frac{1}{2}[/tex] mv²

or,

v = [tex]\sqrt{\frac{2E_k}{m_e} }[/tex]

By substituting the values,

= [tex]\sqrt{\frac{2[6500(1.6\times 10^{-19})]}{9.1\times 10^{-31}} }[/tex]

= 4.78 × 10⁷ m/s

hence,

The electric field strength be:

→ E = vB

= 4.78 × 10⁷ × 0.03

= 1.43 × 10⁶ N/C

Thus the above answer is correct.

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