Question

A helium nucleus makes a full rotation in a circle of radius 0.8 m in two seconds. The value of the magnetic field B at the center of the circle will be
A. \[\dfrac{{{{10}^{ - 19}}}}{{{\mu _0}}}\]
B. ${10^{ - 19}}{\mu _0}$
C. $2 \times {10^{ - 19}}{\mu _0}$
D. $\dfrac{{2 \times {{10}^{ - 19}}}}{{{\mu _0}}}$

Answer 1

Hint: We first start by writing the equation of the magnetic field that is $B = \dfrac{{{\mu _0}I}}{{2r}}$. Now for solving this, we need to find the value of $I$, for that, we use the equation \[I = \dfrac{{e\omega }}{{2\pi }}\] then using the values of $r$,$\omega $, and $e$ that are given in the question we can easily find the answer to the question.

Complete Step-by-Step solution:
Here we are asked to find the magnetic field $B$ at the center of the circle as shown in figure 1.

Figure 1

 So we know that the equation of the magnetic field is given by
$B = \dfrac{{{\mu _0}I}}{{2r}}$------------------------------------------- (1)
Here ${\mu _0}$ is the permeability of free space
$I$ is the current due to the revolution of the electrons
$r = 0.8m$ is the radius of the circle given in the question
For solving equation (1) we need to know the value of $I$
We know that a current $I$ is always flowing in the nth orbit of an atom due to its electrons revolution, and the equation corresponding to that is given by
\[I = \dfrac{{e\omega }}{{2\pi }}\]--------------------------------------------- (2)
Here, $e = 2 \times 1.6 \times {10^{ - 19}}C$ is the charge of a He atom.
          $\omega $ is the angular frequency of the nth orbit.
Given that time is equal to 2 second, so from this we can get the frequency as $f = \dfrac{1}{t} = \dfrac{1}{2}Hz$
Now we can also write $f = \dfrac{\omega }{{2\pi }} = \dfrac{1}{2}$------------------------- (3)
Substituting equation (3) and value of $e$ in (2) we get
For\[I = \dfrac{{e\omega }}{{2\pi }} = \dfrac{{2 \times 1.6 \times {{10}^{ - 19}}}}{2} = 1.6 \times {10^{ - 19}}A\]----------------------- (4)
Now substituting equation (4) in equation (1) we get
$B = \dfrac{{{\mu _0}1.6 \times {{10}^{ - 19}}}}{{2 \times 0.8}} = {10^{ - 19}}{\mu _0}$
Hence option B is correct.

Note: For this type of question we first write the formula for the required value for example here we wrote the formula for the magnetic field $B$. Then we have to substitute the values for the variable and if the value for some variable is unknown we need to calculate that too for example here we have to calculate for $I$. After that, we will get the required result.