Question

What is the magnetic field at the position of the dot in the figure (Figure 1) ? Give your answer as a vector.Express your answer in terms of the unit vectors i^, j^, and k^. Use the 'unit vector' button to denote unit vectors in your answer.

What is the magnetic field at the position of the dot in the figure (Figure 1) ? Give your answer as a vector.Express your answer in terms of the unit vectors i cap, j cap, and k cap. Use the 'unit vector' button to denote unit vectors in your answer.

Answer 1

Concepts and reason

The concept required to solve the question is magnetic field at a point.

Initially find the position of the dot and find the cross product of it with the velocity. Then substitute it in the vector form of the expression of magnetic field at a point of interest.

Fundamentals

The expression of the magnetic field at a point of interest is,

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

Here, $\vec B$ is the magnetic field at a point of interest, q is the charge, $\vec v$ is the velocity, r is the distance of the point of interest from the origin, $\vec r$ is the vector form of the distance of point from the origin, $\hat r$ is the unit vector for the distance of the point of interest from origin, and ${\mu _0}$ is the permeability of the free space.

The cross product of velocity $\vec v$ and the distance $\vec r$ is,

$\vec v \times \vec r$

Substitute $\left( {2.0 \times {{10}^7}\;{\rm{m/s}}} \right)\hat j$ for $\vec v$ and $\left[ {\left( { - 0.02\;{\rm{m}}} \right)\hat i + \left( { - 0.02\;{\rm{m}}} \right)\hat j} \right]$ for $\vec r$ in $\vec v \times \vec r$ .

$\begin{array}{c}\\\vec v \times \vec r = \left[ {\left( {2.0 \times {{10}^7}\;{\rm{m/s}}} \right)\hat j} \right] \times \left[ {\left( { - 0.02\;{\rm{m}}} \right)\hat i + \left( { - 0.02\;{\rm{m}}} \right)\hat j} \right]\\\\ = \left( { - 0.04 \times {{10}^7}\;{{\rm{m}}^2}{\rm{/s}}} \right)\left( { - \hat k} \right)\\\\ = \left( {0.04 \times {{10}^7}\;{{\rm{m}}^2}{\rm{/s}}} \right)\left( {\hat k} \right)\\\end{array}$

The magnitude of the vector $\vec r$ is,

$r = \sqrt {r_x^2 + r_y^2}$

Substitute 0.02 m for ${r_x}$ and ${r_y}$ .

$\begin{array}{c}\\r = \sqrt {{{\left( {0.02\;{\rm{m}}} \right)}^2} + {{\left( {0.02\;{\rm{m}}} \right)}^2}} \\\\ = 0.028\;{\rm{m}}\\\end{array}$

The formula of magnetic field at a point of interest is,

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

Substitute $1.6 \times {10^{ - 19}}\;{\rm{C}}$ for q, $\left( {0.04 \times {{10}^7}\;{{\rm{m}}^2}{\rm{/s}}} \right)\left( {\hat k} \right)$ for $\vec v \times \vec r$ , $4\pi \times {10^7}\;{\rm{Wb/A}}$ for ${\mu _0}$ , $\hat r$ is the unit vector direction, and $\left( {0.028\;{\rm{m}}} \right)$ for r.

$\begin{array}{c}\\\vec B = \frac{{4\pi \times {{10}^7}\;{\rm{Wb/A}}}}{{4\pi }}\frac{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {0.04 \times {{10}^7}\;{{\rm{m}}^2}{\rm{/s}}} \right)\left( {\hat k} \right)}}{{{{\left( {0.028\;{\rm{m}}} \right)}^3}}}\\\\\vec B = \left[ {2.83 \times {{10}^{ - 16}}\left( {\hat k} \right)} \right]\;{\rm{T}}\\\end{array}$

Ans:

The magnetic field at the position of the dot is $\vec B = \left[ {2.83 \times {{10}^{ - 16}}\left( {\hat k} \right)} \right]\;{\rm{T}}$

Answer 2

The answer above is good work, they just wrote the permeability constant wrong in the final equation.