Question

A helium nucleus is moving in a circular path of radius 0.8 m. If it takes 2 sec to complete one revolution. Find out magnetic field produced at the centre of the circle:
A) ${\mu _o} \times {10^{ - 19}}T$
B) $\dfrac{{{{10}^{ - 19}}}}{{{\mu _o}}}T$
C) $2 \times {10^{19}}T$
D) $\dfrac{{2 \times {{10}^{ - 19}}}}{{{\mu _o}}}T$

Answer 1

Hint:Current is defined as the rate of change of charge with respect to time. The unit of current is amperes the unit of charge is coulombs. A moving charge always produces a magnetic field. The unit of magnetic field is Tesla.

Formula used: The formula of the magnetic field produced in a circular coil having current is given by,
$B = \dfrac{{{\mu _o}\left( {N \cdot I} \right)}}{{2R}}$
Where I is the current in the coil N is the number of turns R is the radius and B is the magnetic field produced.
The formula of current is given by,
$I = \dfrac{q}{t}$
Where I is the current in amperes q is the charge in coulombs and t is the time taken in sec.

Complete step by step answer:
It is given in the problem that a helium nucleus is moving in a circular path of radius 0.8 m if it takes 2 sec to complete one revolution then we need to find the value of the magnetic field which is produced by the moving charge in the centre of the circular track in which the charge in revolving.
The current in the coil can be calculated as follows,
The formula of current is given by,
$I = \dfrac{q}{t}$
Where I is the current in amperes q is the charge in coulombs and t is the time taken in sec.
$I = \dfrac{{2e}}{t}$
$I = \dfrac{{2 \times \left( {1 \cdot 6 \times {{10}^{ - 19}}} \right)}}{2}$
$I = 1 \cdot 6 \times {10^{ - 19}}A$
The magnetic field produced in a circular coil having current is given by,
$B = \dfrac{{{\mu _o}\left( {N \cdot I} \right)}}{{2R}}$
Where I is the current in the coil N is the number of turns R is the radius and B is the magnetic field produced.
$ \Rightarrow B = \dfrac{{{\mu _o}\left( {N \cdot I} \right)}}{{2R}}$
Replace the value of current, the number of turns that the charge makes and radius of the circle.
$ \Rightarrow B = \dfrac{{\left( {{\mu _o}} \right) \times \left( {1 \times 1 \cdot 6 \times {{10}^{ - 19}}} \right)}}{{\left( {2 \times 0 \cdot 8} \right)}}$
$ \Rightarrow B = {\mu _o} \times {10^{ - 19}}T$
The magnetic field produced will be$B = {\mu _o} \times {10^{ - 19}}T$.

The correct option for this problem A.

Note:The Helium atom has an atomic number 2 which means that it has two electrons in its shell orbiting the nucleus and also two protons inside the nucleus as the atomic has to be electrically neutral. The charge of the electron is always fixed and equal to $e = 1 \cdot 6 \times {10^{ - 19}}C$.