Question

Radio astronomers detect electromagnetic radiation at 44.0 MHz from an interstellar gas cloud. They suspect this radiation is emitted by electrons spiraling in amagnetic field.

What is the magnetic field strength inside the gas cloud?

Answer 1

Concepts and reason

The concept used to solve this problem is the expression of magnetic field strength in term of frequency and charge.

First write the expression of magnetic field in term of frequency, mass and charge, after that calculate the strength of magnetic field by substitute all given value.

Fundamentals

The electron is moving in circular motion then due to this the expression for strength of magnetic field is,

$B = \frac{{2\pi fm}}{q}$

Here, $B$ is strength of magnetic field, $f$ is frequency, $m$ is mass, and $q$ is charge.

Detected electromagnetic frequency from electromagnetic radiation is,

$\begin{array}{c}\\f = 44.0{\rm{ MHz}}\\\\ = \left( {44.0{\rm{ MHz}}} \right)\left( {\frac{{{{10}^6}{\rm{ Hz}}}}{{1{\rm{ MHz}}}}} \right)\\\\ = \left( {44.0 \times {{10}^6}{\rm{ Hz}}} \right)\\\end{array}$

Mass of electron is,

$m = 9.11 \times {10^{ - 31}}{\rm{ kg}}$

And the charge on electron is,

$q = 1.6 \times {10^{ - 19}}{\rm{ C}}$

The expression for strength of magnetic field is,

$B = \frac{{2\pi fm}}{q}$

Here, $B$ is strength of magnetic field, $f$ is frequency, $m$ is mass, and $q$ is charge.

Substitute $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $q$, $9.11 \times {10^{ - 31}}{\rm{ kg}}$ for $m$, and $44.0 \times {10^6}{\rm{ Hz}}$ for $f$.

$\begin{array}{c}\\B = \frac{{2\pi \left( {44.0 \times {{10}^6}{\rm{ Hz}}} \right)\left( {9.11 \times {{10}^{ - 31}}{\rm{ kg}}} \right)}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\ = 1.57 \times {10^{ - 3}}{\rm{ T}}\\\end{array}$

Ans:

Hence the magnetic field strength inside the gas cloud is$1.57 \times {10^{ - 3}}{\rm{ T}}$.

Answer 2

Frequency f = 44 MHz = 44x10⁶ Hz
Mass (m) of the electrons = 9.11x10^-31 kg
Charge (q) of the electron = 1.6x10^-19 C
When the electrons are moving in circular motion,the magnetic fieldisB =(2πf)(m)/ (q)substituting the given data in above equation, we getB = 1.57*10^-3 TB = 1.57 mT