At time t₁, an electron is sent along the positive direction of an x-axis, through both an electric fieldand a magnetic field$\mathbf{}\overrightarrow{\mathbf{B}}$, with$\overrightarrow{\mathbf{E}}$directed parallel to the y-axis. Figure 28-33 gives the ycomponent F_{net, y}of the net force on the electron due to the two fields, as a function of theelectron’s speed vat time t₁.The scale of the velocity axis is set by${\mathbf{v}}_x\mathbf{=}$100.0 m/s. The xand zcomponents of the net force are zero at t₁. Assuming${\mathbf{B}}_x\mathbf{=}\text{0}$, find

(a)the magnitude E and

(b $\overrightarrow{\mathbf{B}}\mathbf{}$)in unit-vector notation.

When,$v=0,$$F=-{\text{2×10}}^{\text{-19}}\text{N}$

The direction of the magnetic forceisperpendicular to the plane formed by$\overline{\mathbf{v}}$and$\overline{\mathbf{B}}$as determined by the right-hand rule.

Right Hand Rule states that if we arrange our thumb, forefinger, and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the motion of the conductor relative to the magnetic field, and the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

Formulae are as follows:

role="math" localid="1663013317461" $$\begin{gathered} \text{E}=\frac{\text{F}}{\text{q}} \\ =\text{q}(\stackrel{\rightharpoonup }{V}\times \stackrel{\rightharpoonup }{B}) \end{gathered}$$

Where F is a magnetic force, v is velocity, E is the electric field, B is the magnetic field, and q is the charge of the particle.

To find the magnitude of E:

Here,

$$\begin{gathered} \text{F}=\text{qE} \\ \begin{array}{c}E=\frac{F}{q}\\ =\frac{-2\times {10}^{-19}N}{-1.6\times {10}^{-19}C}\\ =1.25N/C\end{array} \end{gathered}$$

Hence, the magnitude of E is$1.25N/C$

To find a magnetic field$\left(\stackrel{\rightharpoonup }{B}\right)$:

$\begin{array}{c}B=\frac{E}{v}\\ =\frac{1.25N/C}{50m/s}\\ =0.025T\end{array}$

To find the direction ofrole="math" localid="1663013448881" $\stackrel{\rightharpoonup }{B}$,

$\stackrel{\rightharpoonup }{F}=\text{q}(\stackrel{\rightharpoonup }{V}\times \stackrel{\rightharpoonup }{B})$

The net force is directed indirection and velocity in$\text{+x}$direction, so by applying the right-hand rule,$\stackrel{\rightharpoonup }{B}$must be directed in$\text{+z}$direction.

Hence,

role="math" localid="1663013638135" $\stackrel{\rightharpoonup }{B}=\text{0.025T}\stackrel{\wedge }{K}$

Hence, the magnetic field is$\stackrel{\rightharpoonup }{B}=\text{0.025T}\stackrel{\wedge }{K}$.

Therefore, the magnitude of the electric field and magnetic field can be determined by using the respective formulae. The direction of the magnetic field can be found by using the right-hand rule.