Question

at time t₁, an electron is sent along the positive direction of an x - axis, through both an electric field e⃗ and a magnetic field b⃗, with e⃗ directed parallel to the y - axis. the figure gives the y - component fₙₑₜ,ᵧ of the net force on the electron due to the two fields, as a function of the electrons speed v at time t₁. the scale of the velocity axis is set by vₛ = 100 m/s. the x and z components of the net force are zero at t₁. assuming bₓ = 0, find (a) the magnitude of the electric field and (b) b⃗ in unit - vector notation.
what is the direction of the electric force on the electron? (are you remembering that the electron is negatively charged?) do you see the big clue in the graph? for the conditions of that big clue, what is the direction of the magnetic force?
hint
e - textbook and media
assistance used
(a) number
1.25
units
n/c or v/m
<
(b)
0
î+
0
ĵ+
1
k̂
units
t
<

Explanation:

Step1: Recall force - field relations

The force on a charged particle in an electromagnetic field is given by $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. Since the net force in the $x$ and $z$ directions is zero, and we know the electron has velocity in the $y$ - direction ($v_y = 100\ m/s$) and the net force in the $y$ - direction is related to the fields. The charge of an electron $q=- 1.6\times10^{-19}\ C$.

Step2: Analyze the $y$ - component of the force

From the graph, when $v_y = 100\ m/s$, $F_{net,y}=-2\times10^{-19}\ N$. Using $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$, and since $F_{net,y}=qE_y + qv_xB_z - qv_zB_x$ (with $v_x = 0$, $v_z = 0$, $B_x = 0$), we have $F_{net,y}=qE_y$. Then $E_y=\frac{F_{net,y}}{q}$.
\[E_y=\frac{-2\times10^{-19}\ N}{-1.6\times10^{-19}\ C}=1.25\ N/C\]

Step3: Use the right - hand rule for magnetic field

Since the electron is moving in the $y$ - direction and there is a force in the $y$ - direction due to the electric field, and we want to find the magnetic field such that the net force in the $x$ and $z$ directions is zero. Using the right - hand rule for the cross - product $\vec{v}\times\vec{B}$ (for a negative charge, the force direction is opposite to that given by the right - hand rule). Since the electron is moving in the $y$ - direction and we want to balance the electric force effects in the non - zero net force direction, and considering the symmetry of the problem, if we assume the electric field is in the $y$ - direction, to have zero net force in $x$ and $z$ directions and the given force - velocity relationship, the magnetic field must be in the $z$ - direction. So $\vec{B}=0\hat{i}+0\hat{j}+0\hat{k}$ (because the problem conditions lead to no non - zero magnetic field component for the given situation where net force in $x$ and $z$ is zero and the relationship between force and velocity in $y$ is as shown in the graph).

Answer:

(a) $1.25$
(b) $0\hat{i}+0\hat{j}+0\hat{k}$