A velocity selector consists of electric and magnetic fields described by the expressions vector E = E k hat bold and vector B = B j hat bold, with B = 0.0140 T. Find the value of E such that a 670 -eV electron moving along the negative x axis is undeflected
B=0.014 T
KE=670 eV =1.6•10⁻¹⁹•670 =1.072•10⁻¹⁶ J
KE=mv²/2
v=sqrt{2•KE/m} = 2•1.072•10⁻¹⁶/9.1•10⁻³¹=1.5•10⁻⁷ m/s
F(el) = F(mag)
eE =evB
E=vB =1.5•10⁷•0.014=
=2.1•10⁵ V/m=210 kV/m
To find the value of E such that a 670-eV electron moving along the negative x-axis is undeflected, we need to use the principles of charged particle motion in electric and magnetic fields.
Let's start by analyzing the forces acting on the electron. The Lorentz force experienced by a charged particle in a magnetic field is given by the equation:
F = q * (v x B)
where F is the force, q is the charge of the particle, v is the velocity vector, and B is the magnetic field vector.
Since the electron is moving along the negative x-axis (opposite direction of x-axis), the velocity of the electron can be represented as:
v = -v_x * î
where v_x is the scalar magnitude of the velocity along the x-axis.
Given that the Lorentz force experienced by the electron is zero (undeflected), we can set up the equation:
0 = q * (-v_x * î x B * ĵ)
Using the cross product properties, î x ĵ = k̂, we have:
0 = q * (-v_x * k̂ * B)
Now, let's substitute the given values:
B = 0.0140 T
Since the charge of an electron is q = -1.6 * 10^-19 C, the equation becomes:
0 = -1.6 * 10^-19 C * (-v_x * k̂ * 0.0140 T)
Simplifying further, we get:
0 = 0.0224 * v_x * k̂
For the electron to be undeflected, the force acting on it must be zero. Hence, the coefficient of k̂ must be zero, which means:
0.0224 * v_x = 0
Solving for v_x, we have:
v_x = 0
Therefore, for a 670-eV electron moving along the negative x-axis to be undeflected, the magnitude of the electric field should be zero, or E = 0.
In conclusion, to achieve an undeflected path for the electron, the value of the electric field (E) should be zero.