Chapter 22, Problem 33P

One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of −2.00 μC is moving with a velocity of $150\hat{i}\text{Mm/s}$ along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undetected. Calculate the required vector electric field.

To determine

The location where total magnetic field is zero.

Answer to Problem 33P

The location where total magnetic field is zero is $0.42\text{m}$.

Explanation of Solution

Given info: Current flowing through the first wire is $30.0\text{A}$, current flowing through second wire is $50.0\text{A}$, distance between plane of both wire is $0.28\text{m}$,

Formula to calculate magnetic field due to first wire is,

$B_1=\frac{{\mu }_o{I}_1}{2\pi r_1}$

Here,

${\mu }_o$ is permeability in free space.

$I_1$ is current in first wire.

$r_1$ is is radius of first wire.

Formula to calculate magnetic field due to second wire is,

$B_2=\frac{{\mu }_o{I}_2}{2\pi r_2}$

Here,

$I_2$ is current in second wire.

$r_2$ is radius of second wire.

Total magnetic field given by both wire is,

$B=B_1+B_2$ (1)

Substitute $\frac{{\mu }_o{I}_1}{2\pi r_1}$ for $B_1$ and $\frac{{\mu }_o{I}_2}{2\pi r_2}$ for $B_2$ in equation (1).

$B=\frac{{\mu }_o{I}_1}{2\pi r_1}+\frac{{\mu }_o{I}_2}{2\pi r_2}$ (2)

Substitute $0$ for $B$ as per given condition in part (a),

$\begin{array}{c}0=\frac{{\mu }_o{I}_1}{2\pi r_1}+\frac{{\mu }_o{I}_2}{2\pi r_2}\\ \frac{I_1}{r_1}=\frac{I_2}{r_2}\end{array}$

Substitute $30\text{A}$ for $I_1$, $50\text{A}$ for $I_2$, $\left|y\right|$ for $r_1$ and $\left|y\right|+0.28\text{m}$ for $r_2$ in above equation.

$\begin{array}{c}\frac{30\text{A}}{\left|y\right|}=\frac{50\text{A}}{\left|y\right|+0.28\text{m}}\\ 50\text{A}\left(\left|y\right|\right)=30\text{A}\left(\left|y\right|+0.28\text{m}\right)\\ 20\left(\left|y\right|\right)=30\times 0.28\\ \left|y\right|=0.42\text{m}\end{array}$

Thus, the location where total magnetic field is zero is $0.42\text{m}$.

Conclusion:

Therefore, the location where total magnetic field is zero is $0.42\text{m}$.

To determine

The magnitude of vector magnetic force.

Answer to Problem 33P

The magnitude of vector magnetic force is $3.47\times {10}^{-2}\left(\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

Given info: charge on particle on particle is $-2\text{μC}$, velocity of particle is $150\widehat{\text{i}}\text{Mm}/\text{s}$, location of particle is $0.1\text{m}$.

Formula to calculate total magnetic field is,

$\begin{array}{c}B=\frac{{\mu }_o{I}_1}{2\pi r_1}+\frac{{\mu }_o{I}_2}{2\pi r_2}\\ =\frac{{\mu }_o{I}_1}{2\pi \left|y\right|}+\frac{{\mu }_o{I}_2}{2\pi \left(\left|y\right|+0.28\text{m}\right)}\\ =\frac{{\mu }_o}{2\pi }\left(\frac{I_1}{\left|y\right|}+\frac{I_2}{\left(\left|y\right|+0.28\text{m}\right)}\right)\end{array}$

Substitute $4\pi \times {10}^{-7}\text{Tm}/\text{A}$ for ${\mu }_o$, $30\text{}\text{A}$ for $I_1$, $50\text{A}$ for and $0.1\text{m}$ for $\left|y\right|$ in above equation.

$\begin{array}{c}B=\frac{4\pi \times {10}^{-7}\text{Tm}/\text{A}}{2\pi }\left(\frac{30\text{A}}{0.1\text{m}}\left(-\widehat{\text{k}}\right)+\frac{50\text{A}}{\left(0.28-0.1\right)\text{m}}\left(-\widehat{\text{k}}\right)\right)\\ =\left(1.16\times {10}^{-4}-\widehat{\text{k}}\right)\text{T}\end{array}$

Formula to calculate force acting on particle is,

$F=qvB$

Here,

$q$ is charge on particle.

$v$ is the velocity of particle.

$B$ is magnetic field due to wire.

Substitute $1.16\times {10}^{-4}\text{}\text{T}\left(-\stackrel{\wedge }{k}\right)$ for $B$, $-2\text{μC}$ for $q$ and $150\widehat{\text{i}}\text{Mm}/\text{s}$ for $v$ in above equation.

$\begin{array}{c}F=\left(-2\text{μC}\right)\left(150\widehat{\text{i}}\text{Mm}/\text{s}\right)\left(1.16\times {10}^{-4}\text{T}\left(-\widehat{\text{k}}\right)\right)\\ =\left(-2\text{μC}\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\left(150\widehat{\text{i}}\text{Mm}/\text{s}\frac{{10}^6\text{m}/\text{s}}{1\text{Mm}/\text{s}}\right)\left(1.16\times {10}^{-4}\text{T}\left(-\widehat{\text{k}}\right)\right)\\ =3.47\times {10}^{-2}\text{N}\left(-\widehat{\text{i}}\widehat{\text{k}}\right)\\ =3.47\times {10}^{-2}\left(\widehat{\text{j}}\right)\text{N}\end{array}$

Hence, magnitude of vector magnetic force is $3.47\times {10}^{-2}\left(\widehat{\text{j}}\right)\text{N}$.

Conclusion:

Therefore, magnitude of vector magnetic force is $3.47\times {10}^{-2}\left(\widehat{\text{j}}\right)\text{N}$.

To determine

The magnitude of vector electric field.

Answer to Problem 33P

The magnitude of vector electric field is $-1.73\times {10}^4\text{N}/\text{C}\left(\widehat{\text{j}}\right)$.

Explanation of Solution

Formula to calculate electric field is,

$E=\frac{F}{q}$

Here,

$F$ force acting on charged particle.

$q$ is charge on particle.

Substitute $3.47\times {10}^{-2}\text{N}\left(-\widehat{\text{j}}\right)$ for $F$ and $-2\text{μC}$ for $q$ in above equation.

$\begin{array}{c}E=\frac{3.47\times {10}^{-2}\text{N}\left(\widehat{\text{j}}\right)}{-2\text{μC}}\\ =\frac{3.47\times {10}^{-2}\text{N}\left(\widehat{\text{j}}\right)}{-2\text{μC}\frac{{10}^{-6}\text{C}}{1\text{μC}}}\\ =-1.73\times {10}^4\left(\widehat{\text{j}}\right)\text{N}/\text{C}\end{array}$

Hence, magnitude of vector electric field is $-1.73\times {10}^4\left(\widehat{\text{j}}\right)\text{N}/\text{C}$.

Conclusion:

Therefore, magnitude of vector electric field is $-1.73\times {10}^4\left(\widehat{\text{j}}\right)\text{N}/\text{C}$.