Question

Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.93 ✕ 105 m/s in the positive x-direction. Thousands of miles from Earth, they interact with Earth's magnetic field of magnitude 3.04 ✕ 10−8 T in the positive z-direction. Find the magnitude and direction of the magnetic force on a proton. Find the magnitude and direction of the magnetic force on an electron.

HINT
(a)
magnitude of the magnetic force on a proton (in N)
N
(b)
direction of the magnetic force on a proton
+x-direction
−x-direction
+y-direction
−y-direction
+z-direction
−z-direction
(c)
magnitude of the magnetic force on an electron (in N)
N
(d)
direction of the magnetic force on an electron
+x-direction
−x-direction
+y-direction
−y-direction
+z-direction
−z-direction

Answer 1

(a) Approximately
`1.91 * 10^(-21)\; {\rm N}`.

(b)
`(-y)`-direction.

(c) Approximately
`1.91 * 10^(-21)\; {\rm N}`.

(d)
`(+y)`-direction.

Step-by-step explanation:

The magnetic force on a moving charge can be found with the equation:

`\displaystyle F = q\, \left(\vec{v}* \vec{B}\right)`,

Where:

• 
  `q` is the electrostatic charge on the moving object (
  `q` might be negative, as in the case of the electron,)
• 
  `\vec{v}` is the velocity vector of the moving object, and
• 
  `\vec{B}` is the magnetic field vector.

In this expression,
`\vec{v} * \vec{B}` is the vector cross-product between
`\vec{v}` and
`\vec{B}`. Note the order of this cross-product: the velocity vector comes before the magnetic field vector.

The electrostatic charge on a proton is equal to the elementary charge, a physical constant:

`q \approx 1.602 * 10^(-19)\; {\rm C}`.

In this question, it is given that the speed (a scalar) of the proton is
`v = 3.93 * 10^(5)\; {\rm m\cdot s^(-1)}`. Since this proton is moving in the positive
`x`-direction, the
`x\!`-component (first component) of the velocity vector
`\vec{v}` should be positive, while the other two components should be zero. Hence, the velocity vector of this proton would be:

`\begin{aligned} \vec{v} &= \begin{bmatrix}3.93 * 10^(5) \\ 0 \\ 0\end{bmatrix}\; {\rm m\cdot s^(-1) \\ &= (3.93 * 10^(5))\, \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\; {\rm m\cdot s^(-1)} \end{aligned}`.

Similarly, since the magnetic field points in the positive
`z`-direction, the
`z\!`-component of the magnetic field vector
`\vec{B}` should be positive, while the other two components should be zero:

`\begin{aligned} \vec{B} &= \begin{bmatrix} 0 \\ 0\\ 3.04 * 10^(-8)\end{bmatrix}\; {\rm m\cdot s^(-1) \\ &= (3.04 * 10^(-8))\, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\; {\rm m\cdot s^(-1)} \end{aligned}`.

The magnetic force on this proton would be:

`\begin{aligned} F &= q\, \left(\vec{v}* \vec{B}\right) \\ &\approx (1.602 * 10^(-19))\, (3.93 * 10^(5))\, (3.04 * 10^(-8))\, \left(\begin{bmatrix}1 \\0 \\ 0\end{bmatrix} * \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)\; {\rm N} \\ &\approx (1.91 * 10^(-21)\; {\rm N})\, \begin{bmatrix}0 * 1 - 0 * 0 \\ 0 * 0 - 1 * 1 \\ 1 * 0 - 0 * 0\end{bmatrix} \\ &= (1.91 * 10^(-21)\; {\rm N})\, \begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}\end{aligned}`.

In other words, magnitude of the magnetic force on this proton would be approximately
`1.91 * 10^(21)\; {\rm N}`. Since the
`y`-component of this force is negative while the
`x`- and
`z`-components are zero, this force would point in the
`(-y)`-direction (negative
`y\!` direction.)

The electrostatic charge on an electron is the opposite of that on a proton:

`q \approx (-1.602) * 10^(-19)\; {\rm C}`.

The value of
`\vec{v}` and
`\vec{B}` are the same as the ones in the case of the proton. The magnetic force vector on this electron would be:

`\begin{aligned} F &= q\, \left(\vec{v}* \vec{B}\right) \\ &\approx ((-1.602) * 10^(-19))\, (3.93 * 10^(5))\, (3.04 * 10^(-8))\, \left(\begin{bmatrix}1 \\0 \\ 0\end{bmatrix} * \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)\; {\rm N} \\ &\approx ((-1.91) * 10^(-21)\; {\rm N})\, \begin{bmatrix}0 * 1 - 0 * 0 \\ 0 * 0 - 1 * 1 \\ 1 * 0 - 0 * 0\end{bmatrix} \\ &= (1.91 * 10^(-21)\; {\rm N})\, \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}\end{aligned}`.

The magnitude of the magnetic force on the electron would be
`1.91 * 10^(-21)\; {\rm N}`, same as that on the proton. While the
`x`- and
`z`-components are still zero, this force vector would point in the
`(+y)`-direction (positive
`y\!`-direction) since the
`y`-component is now positive.