Question

A +6.00 -μC point charge is moving at a constant 8.00×106 m/s in the + y-direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic-field vector it produces at the following points.

Part A: x = +.5 m, y = 0 m, z = 0 m

Part B: x = 0 m, y = -.5 m, z = 0 m

Part C: x = 0 m, y = 0 m, z = +.5 m

Part D: x = 0 m, y = -.5 m, z = +.5 m

Answer 1

Concepts and reason

The required concepts to solve these questions are Biot-Savart law, direction of the magnetic field and the general rule of the cross product.

Biot-Savart Law: This law is used to determine the magnitude and direction of the magnetic field produced. Magnetic field is produced by a moving charge.

Calculate the magnetic field by using the Biot-Savart law at different points.

init

Fundamentals

The expression for Biot-Savart law is given by,

$\vec B = \frac{{{\mu _0}}}{{4\pi }}\frac{{q\vec v \times \hat r}}{{{r^2}}}$

Here,$q$is the charge, $r$is the magnitude of the displacement vector, ${\mu _0}$is the permeability and $\hat r$ is the unit vector and$v$ is the velocity.

The expression of the unit vector can be written as,

$\hat r = \frac{{\vec r}}{{\left| {\vec r} \right|}}$

Here, $\hat r$is the unit vector.

(a)

The expression for Biot-Savart law is given by,

$\vec B = \frac{{{\mu _0}}}{{4\pi }}\frac{{q\vec v \times \hat r}}{{{r^2}}}$

Substitute $\frac{{\vec r}}{{\left| {\vec r} \right|}}$ for $\hat r$in the above equation.

$\vec B = \frac{{{\mu _0}}}{{4\pi }}\frac{{q\vec v \times \vec r}}{{{r^3}}}$ …… (1)

At $x = + 0.5{\rm{ m}}$, the displacement vector is,

$\vec r = \left( {0.5{\rm{ m}}} \right)\vec i + 0\vec j + 0\vec k$

The magnitude of displacement vector is,

$\begin{array}{c}\\r = \sqrt {{{\left( {0.5{\rm{ m}}} \right)}^2} + {0^2} + {0^2}} \\\\ = 0.5{\rm{ m }}\\\end{array}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}$ for ${\mu _0}$ , $6{\rm{ \mu C}}$for$q$, $\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j$for $\vec v$, $0.5{\rm{ m }}$ for $r$ and $\left( {0.5{\rm{ m}}} \right)\vec i$ for$\vec r$ in the equation (1).

$\begin{array}{c}\\\vec B = \left( {\frac{{4\pi \times {{10}^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}}}{{4\pi }}} \right)\left( {6{\rm{ \mu C}}} \right)\left( {\frac{{1{\rm{ C}}}}{{{{10}^6}{\rm{ \mu C}}}}} \right)\left( {\frac{{\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j \times \left( {0.5{\rm{ m}}} \right)\vec i}}{{\left( {0.5{\rm{ m}}} \right){{\rm{ }}^3}}}} \right)\\\\ = \left( {{{10}^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}} \right)\left( {6 \times {{10}^{ - 6}}{\rm{ C}}} \right)\left( {\frac{{\left( {4 \times {{10}^6}{\rm{ }}{{\rm{m}}^2} \cdot {{\rm{s}}^{ - 2}}} \right)\left( { - \vec k} \right)}}{{0.125{\rm{ }}{{\rm{m}}^3}}}} \right)\\\\ = 1.92 \times {10^{ - 5}}\,{\rm{T}}\left( { - \vec k} \right)\\\\\, = \left( {0{\rm{ T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j - \left( {1.92 \times {{10}^{ - 5}}\,{\rm{T}}} \right)\vec k\\\end{array}$

(b)

Use equation (1) to calculate the magnetic field at point $y = - 0.5{\rm{ m}}$.

$\vec B = \frac{{{\mu _0}}}{{4\pi }}\frac{{q\vec v \times \vec r}}{{{r^3}}}$

The displacement vector is,

$\vec r = 0\vec i + \left( { - 0.5{\rm{ m}}} \right)\vec j + 0\vec k$

The magnitude of displacement vector is,

$\begin{array}{c}\\r = \sqrt {{0^2} + {{\left( { - 0.5{\rm{ m}}} \right)}^2} + {0^2}} \\\\ = 0.5{\rm{ m}}\\\end{array}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}$ for ${\mu _0}$ , $6{\rm{ \mu C}}$for$q$, $\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j$for $\vec v$, $0.5{\rm{ m }}$ for $r$ and $\left( { - 0.5{\rm{ m}}} \right)\vec j$ for$\vec r$ in the equation (1).

$\begin{array}{c}\\\vec B = \left( {\frac{{4\pi \times {{10}^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}}}{{4\pi }}} \right)\left( {6{\rm{ \mu C}}} \right)\left( {\frac{{1{\rm{ C}}}}{{{{10}^6}{\rm{ \mu C}}}}} \right)\left( {\frac{{\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j \times \left( { - 0.5{\rm{ m}}} \right)\vec j}}{{\left( {0.5{\rm{ m}}} \right){{\rm{ }}^3}}}} \right)\\\\ = \left( {0{\rm{ T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j + 0\vec k\\\end{array}$

(c)

Use equation (1) to calculate the magnetic field at point $z = 0.5{\rm{ m}}$.

$\vec B = \frac{{{\mu _0}}}{{4\pi }}\frac{{q\vec v \times \vec r}}{{{r^3}}}$

The displacement vector is,

$\vec r = 0\vec i + 0\vec j + \left( {0.5{\rm{ m}}} \right)\vec k$

The magnitude of displacement vector is,

$\begin{array}{c}\\r = \sqrt {{0^2} + {0^2} + {{\left( {0.5{\rm{ m}}} \right)}^2}} \\\\ = 0.5{\rm{ m}}\\\end{array}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}$ for ${\mu _0}$ , $6{\rm{ \mu C}}$for$q$, $\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j$for $\vec v$, $0.5{\rm{ m }}$ for $r$ and $\left( {0.5{\rm{ m}}} \right)\vec k$ for$\vec r$ in the equation (1).

$\begin{array}{c}\\\vec B = \left( {\frac{{4\pi \times {{10}^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}}}{{4\pi }}} \right)\left( {6{\rm{ \mu C}}} \right)\left( {\frac{{1{\rm{ C}}}}{{{{10}^6}{\rm{ \mu C}}}}} \right)\left( {\frac{{\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j \times \left( {0.5{\rm{ m}}} \right)\vec k}}{{\left( {0.5{\rm{ m}}} \right){{\rm{ }}^3}}}} \right)\\\\ = \left( {1.92 \times {{10}^{ - 5}}\,{\rm{T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j + \left( {0{\rm{ T}}} \right)\vec k\\\end{array}$

(d)

Use equation (1) to calculate the magnetic field at points $x = 0{\rm{ m}}$,$y = - 0.5{\rm{ m}}$,$z = + 0.5{\rm{ m}}$

$\vec B = \frac{{{\mu _0}}}{{4\pi }}\frac{{q\vec v \times \vec r}}{{{r^3}}}$

The displacement vector is,

$\vec r = 0\vec i + \left( { - 0.5{\rm{ m}}} \right)\vec j + \left( {0.5{\rm{ m}}} \right)\vec k$

The magnitude of displacement vector is,

$\begin{array}{c}\\r = \sqrt {{0^2} + {{\left( {0.5{\rm{ m}}} \right)}^2} + {{\left( {0.5{\rm{ m}}} \right)}^2}} \\\\ = 0.707{\rm{ m}}\\\end{array}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}$ for ${\mu _0}$ , $6{\rm{ \mu C}}$for$q$, $\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j$for $\vec v$, $0.707{\rm{ m}}$ for $r$ and $0\vec i + \left( { - 0.5{\rm{ m}}} \right)\vec j + \left( {0.5{\rm{ m}}} \right)\vec k$ for$\vec r$ in the equation (1).

$\begin{array}{c}\\\vec B = \left( {\frac{{4\pi \times {{10}^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}}}{{4\pi }}} \right)\left( {6{\rm{ \mu C}}} \right)\left( {\frac{{1{\rm{ C}}}}{{{{10}^6}{\rm{ \mu C}}}}} \right)\left( {\frac{{\left( {8 \times {{10}^6}{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\vec j \times \left( {\left( { - 0.5{\rm{ m}}} \right)\vec j + \left( {0.5{\rm{ m}}} \right)\vec k} \right)}}{{{{\left( {0.707{\rm{ m}}} \right)}^3}}}} \right)\\\\ = \left( {{{10}^{ - 7}}{\rm{ N}} \cdot {{\rm{A}}^{ - 2}}} \right)\left( {6 \times {{10}^{ - 6}}{\rm{ C}}} \right)\left( {\frac{{\left( {4 \times {{10}^6}{{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}} \right)\vec i}}{{0.353{\rm{ }}{{\rm{m}}^3}}}} \right)\\\\ = \left( {6.79 \times {{10}^{ - 6}}\,{\rm{T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j + \left( {0{\rm{ T}}} \right)\vec k\\\end{array}$

Ans: Part a

The magnetic field at points $x = + 0.5{\rm{ m}}$, $y = 0{\rm{ m}}$, $z = 0{\rm{ m}}$are $\left( {0{\rm{ T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j - \left( {1.92 \times {{10}^{ - 5}}\,{\rm{T}}} \right)\vec k$.

Part b

The magnetic field at points $x = 0{\rm{ m}}$,$y = - 0.5{\rm{ m}}$,$z = 0{\rm{ m}}$are$\left( {0{\rm{ T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j + 0\vec k$.

Part c

The magnetic field at points $x = 0{\rm{ m}}$,$y = 0$,$z = 0.5{\rm{ m}}$are $\left( {1.92 \times {{10}^{ - 5}}\,{\rm{T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j + \left( {0{\rm{ T}}} \right)\vec k$.

Part d

The magnetic field at points $x = 0{\rm{ m}}$,$y = - 0.5{\rm{ m}}$,$z = + 0.5{\rm{ m}}$are $\left( {6.79 \times {{10}^{ - 6}}\,{\rm{T}}} \right)\vec i + \left( {0{\rm{ T}}} \right)\vec j + \left( {0{\rm{ T}}} \right)\vec k$.