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Expert-verified Found in: Page 829 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # At time t1, an electron is sent along the positive direction of an x-axis, through both an electric field and a magnetic field$\mathbf{}\stackrel{\mathbf{\to }}{\mathbf{B}}$, with $\stackrel{\mathbf{\to }}{\mathbf{E}}$directed parallel to the y-axis. Figure 28-33 gives the y component Fnet, y of the net force on the electron due to the two fields, as a function of theelectron’s speed v at time t1.The scale of the velocity axis is set by ${{\mathbf{v}}}_{x}{\mathbf{=}}$100.0 m/s. The x and z components of the net force are zero at t1. Assuming${{\mathbf{B}}}_{x}{\mathbf{=}}{\text{0}}$ , find (a)the magnitude E and (b $\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{}}$)in unit-vector notation. a.$\text{E}=\text{1.25 V/m}$

b.$\stackrel{⇀}{B}=0.\text{025 T}\stackrel{\wedge }{K}$

See the step by step solution

## Step 1: Given

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

## Step 2: Determining the concept

The direction of the magnetic force is perpendicular 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" $\text{E}=\frac{\text{F}}{\text{q}}\phantom{\rule{0ex}{0ex}}=\text{q}\left(\stackrel{⇀}{V}×\stackrel{⇀}{B}\right)$

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.

## Step 3: (a) Determining the magnitude

To find the magnitude of E:

Here,

$\text{F}=\text{qE}\phantom{\rule{0ex}{0ex}}\begin{array}{c}E=\frac{F}{q}\\ =\frac{-2×{10}^{-19}\text{\hspace{0.17em}}N}{-1.6×{10}^{-19}\text{\hspace{0.17em}}C}\\ =1.25\text{\hspace{0.17em}}N/C\end{array}$

Hence, the magnitude of E is $1.25\text{\hspace{0.17em}}N/C$

## Step 4: (b) Determining the B→ in-unit vector notation

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

$\begin{array}{c}B=\frac{E}{v}\\ =\frac{1.25\text{\hspace{0.17em}}N/C}{50\text{\hspace{0.17em}}m/s}\\ =0.025\text{\hspace{0.17em}}T\end{array}$

To find the direction of role="math" localid="1663013448881" $\stackrel{⇀}{B}$,

$\stackrel{⇀}{F}=\text{q}\left(\stackrel{⇀}{V}×\stackrel{⇀}{B}\right)$

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

Hence,

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

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

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. ### Want to see more solutions like these? 