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
Well, it seems like we have an electron that's trying to make its way through this velocity selector without any distractions. Hang on tight, because we're about to mathematically tickle this problem!
First things first, let's write down the Lorentz force equation for this electron:
F = q(E + v × B)
Since the electron is undeflected, the net force on it must be zero. So we have:
0 = q(E + v × B)
Here, q is the charge of the electron, which is 1.6 x 10^-19 C, and v is the velocity vector of the electron. Given that the velocity of the electron is along the negative x-axis, we have v = -v₀ î, where v₀ is the magnitude of the velocity.
Plugging in these values, we get:
0 = (1.6 x 10^-19 C)(E - v₀ B ĵ)
Now, let's look at the y-component of this equation. Since the electron is moving along the negative x-axis, v₀ is negative, so -v₀ B ĵ becomes a positive value.
0 = E(1.6 x 10^-19 C) - (670 eV)(1.6 x 10^-19 C)(0.0140 T)
Simplifying this equation, we get:
E = (670 eV)(0.0140 T)
And finally, computing this...
E = 9.38 V
So, the value of E that keeps our electron undistracted is approximately 9.38 V! Just the right amount of electric field to keep this electron focused and going straight like an arrow.
To find the value of E such that the electron is undeflected, we need to consider the forces acting on the electron.
The electric force (F_E) on the electron is given by the equation F_E = qE, where q is the charge of the electron. Since the electron has a negative charge, q = -1.602 x 10^-19 C.
The magnetic force (F_B) on the electron is given by the equation F_B = qvB, where v is the velocity of the electron. In this case, the electron is moving along the negative x-axis, so v = -v_0, where v_0 is the speed of the electron.
For the electron to be undeflected, the electric force and the magnetic force must be equal in magnitude and opposite in direction.
Equating the forces, we have:
F_E = F_B
qE = qvB
E = vB
Now we can substitute the given values:
B = 0.0140 T
v = -v_0
Therefore,
E = (-v_0)(0.0140 T)
Since we are given the energy of the electron as 670 eV, we can use the equation for kinetic energy:
K.E = (1/2)mv^2
where m is the mass of the electron. The kinetic energy can be expressed in terms of the electron's speed:
K.E = (1/2)mv_0^2
Setting K.E equal to 670 eV and using the conversion factor 1 eV = 1.602 x 10^-19 J, we have:
(1/2)mv_0^2 = 670 eV * (1.602 x 10^-19 J/eV)
= 1.0734 x 10^-16 J
Now we can solve for v_0:
v_0^2 = (2 * 1.0734 x 10^-16 J) / m
From relativistic physics, the relativistic kinetic energy can be expressed as:
K.E = mc^2 / (sqrt(1 - v^2/c^2)) - mc^2
where c is the speed of light.
Substituting the given energy, we have:
1.0734 x 10^-16 J = mc^2 / (sqrt(1 - v_0^2/c^2)) - mc^2
Simplifying, we get:
sqrt(1 - v_0^2/c^2) = mc^2 / (mc^2 + 1.0734 x 10^-16 J)
Now we can solve for v_0/c:
1 - v_0^2/c^2 = (mc^2)^2 / [(mc^2 + 1.0734 x 10^-16 J)^2]
Simplifying further, we get:
v_0^2/c^2 = 1 - [(mc^2)^2 / (mc^2 + 1.0734 x 10^-16 J)^2]
Finally, we can solve for v_0:
v_0 = c * sqrt[1 - [(mc^2)^2 / (mc^2 + 1.0734 x 10^-16 J)^2]]
Now that we know v_0, we can substitute it into the equation for E:
E = (-v_0)(0.0140 T)
Simplifying the expression will give us the value of E.
To find the value of E such that the electron is undeflected, we need to set the electric force equal to the magnetic force acting on the electron.
The electric force on a charged particle is given by the equation F_electric = q * E, where q is the charge of the particle and E is the electric field.
The magnetic force on a charged particle is given by the equation F_magnetic = q * v * B, where q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.
In this case, we have an electron moving along the negative x-axis, so the velocity vector can be written as v = -v_0 i-hat, where v_0 is the magnitude of the velocity.
Setting the electric force equal to the magnetic force, we have:
q * E = q * (-v_0) * B
Since the charge of an electron is q = -1.6 x 10^-19 C, we can substitute this value in the equation:
(-1.6 x 10^-19 C) * E = (-1.6 x 10^-19 C) * (-v_0) * (0.0140 T)
The charge of the electron cancels out on both sides of the equation:
E = v_0 * (0.0140 T)
To find the value of E, we need to determine the magnitude of the velocity (v_0) of the electron.
Since we know that the electron has an energy of 670 eV, we can use the equation for kinetic energy:
K.E. = (1/2) * m * v^2
Where K.E. is the kinetic energy, m is the mass of the electron, and v is the velocity.
Converting the given energy of 670 eV to joules:
670 eV * (1.6 x 10^-19 J/eV) = 1.072 x 10^-16 J
For an electron, the mass is given by m = 9.1 x 10^-31 kg.
Plugging the values into the kinetic energy equation, we can solve for the velocity:
1.072 x 10^-16 J = (1/2) * (9.1 x 10^-31 kg) * v^2
v^2 = (2 * 1.072 x 10^-16 J) / (9.1 x 10^-31 kg)
v^2 = 2.35 x 10^14 m^2/s^2
Taking the square root of both sides:
v = 4.85 x 10^7 m/s
Substituting this value of v_0 into the equation for E:
E = (4.85 x 10^7 m/s) * (0.0140 T)
E = 6.79 x 10^5 N/C
Therefore, the value of E that makes the electron undeflected is 6.79 x 10^5 N/C.