
An electron moves in a circular orbit with uniform speed $v$.It produces a magnetic field $B$at the centre of the circle. The radius of the circle is proportional to
A.$\dfrac{B}{v}$
B.$\dfrac{v}{R}$
C.$\sqrt{\dfrac{v}{B}}$
D.$\sqrt{\dfrac{B}{v}}$
Answer
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Hint: When an electron moves in a circular orbit with uniform speed producing a magnetic field at the centre of the circle. This is because a circular current is generated at the centre. Therefore by substituting the value of the equivalent current produced in a circular orbit in Biot-Savart law we can determine the radius of the circle.
Complete answer:
The electron moves in a circular orbit with uniform speed $v$and the magnetic field$B$ is produced at the centre of the circle. The time period of the electron that is moving in an angular orbit can be expressed as:
Time period,$T=\dfrac{2\pi r}{v}$ ……….(i)
Here $r$is the radius of the circle and $v$ the speed of the charged particle.
[since$2\pi r=$ Circumference of the circular path.]
Now the equivalent current,$i$ which is produced in a circle due to the flow of electrons is given by
$i=\dfrac{ch\arg e}{time}=\dfrac{e}{T}$ ………(ii)
Now putting the value $T$in equation (ii),
$i=\dfrac{e}{\dfrac{2\pi r}{v}}$
Or,$i=\dfrac{ev}{2\pi r}$ ……….(iii)
According to Biot-Savart law, the magnetic field $B$at a particular point induced by the flow of current can be expressed as,

$B=\dfrac{{{\mu }_{o}}i}{2r}$ ………(iv)
Here ${{\mu }_{o}}$denotes the permeability constant and $i$is equivalent current produced in a circular path.
Hence by substituting the value of the equivalent current,$i$in equation (iv) we get,
$B=\dfrac{{{\mu }_{o}}}{2r}\times \dfrac{ev}{2\pi r}$
Or,${{r}^{2}}=\dfrac{{{\mu }_{o}}ev}{4\pi B}$
Or,$r=\sqrt{\dfrac{{{\mu }_{o}}ev}{4\pi B}}$
Therefore $r$ $\alpha $ $\sqrt{\dfrac{v}{B}}$
The radius of the circle,$r$, is directly proportional to $\sqrt{\dfrac{v}{B}}$.
Thus, option (C) is correct.
Note: Magnetic permeability is used to determine materials resistance to a magnetic field. The higher the permeability, the higher will be the conductivity of magnetic lines of force. It is found that the materials like iron alloys, nickel alloys, and cobalt alloys have relatively high permeability.
Complete answer:
The electron moves in a circular orbit with uniform speed $v$and the magnetic field$B$ is produced at the centre of the circle. The time period of the electron that is moving in an angular orbit can be expressed as:
Time period,$T=\dfrac{2\pi r}{v}$ ……….(i)
Here $r$is the radius of the circle and $v$ the speed of the charged particle.
[since$2\pi r=$ Circumference of the circular path.]
Now the equivalent current,$i$ which is produced in a circle due to the flow of electrons is given by
$i=\dfrac{ch\arg e}{time}=\dfrac{e}{T}$ ………(ii)
Now putting the value $T$in equation (ii),
$i=\dfrac{e}{\dfrac{2\pi r}{v}}$
Or,$i=\dfrac{ev}{2\pi r}$ ……….(iii)
According to Biot-Savart law, the magnetic field $B$at a particular point induced by the flow of current can be expressed as,

$B=\dfrac{{{\mu }_{o}}i}{2r}$ ………(iv)
Here ${{\mu }_{o}}$denotes the permeability constant and $i$is equivalent current produced in a circular path.
Hence by substituting the value of the equivalent current,$i$in equation (iv) we get,
$B=\dfrac{{{\mu }_{o}}}{2r}\times \dfrac{ev}{2\pi r}$
Or,${{r}^{2}}=\dfrac{{{\mu }_{o}}ev}{4\pi B}$
Or,$r=\sqrt{\dfrac{{{\mu }_{o}}ev}{4\pi B}}$
Therefore $r$ $\alpha $ $\sqrt{\dfrac{v}{B}}$
The radius of the circle,$r$, is directly proportional to $\sqrt{\dfrac{v}{B}}$.
Thus, option (C) is correct.
Note: Magnetic permeability is used to determine materials resistance to a magnetic field. The higher the permeability, the higher will be the conductivity of magnetic lines of force. It is found that the materials like iron alloys, nickel alloys, and cobalt alloys have relatively high permeability.
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