Question

Ratio of magnetic field at the centre of a current carrying coil of radius ${\text{R}}$ and at a distance of ${\text{3 R}}$ on its axis is
$
  {\text{(A) 10}}\sqrt {{\text{10}}} \\
  {\text{(B) 20}}\sqrt {{\text{10}}} \\
  {\text{(C) 2}}\sqrt {{\text{10}}} \\
  {\text{(D) }}\sqrt {{\text{10}}} \\
$

Answer 1

Hint:For finding the ratio of magnetic field at the centre of a current carrying coil of radius ${\text{R}}$ and at some distance ${\text{x}}$ on its axis is given by formula

\[\dfrac{{{{\text{B}}_{{\text{centre}}}}}}{{{{\text{B}}_{{\text{axis}}}}}}{\text{ = }}{\left( {{\text{1 + }}\dfrac{{{x^2}}}{{{{\text{R}}^{\text{2}}}}}} \right)^{{\text{3/2}}}}\] and then substitute the values of ${\text{x}}$ and ${\text{R}}$ in formula and simplify.

Complete step by step solution: Given: Radius of first current carrying coil is ${\text{R}}$
Distance, \[{\text{x = 3R}}\]

Magnetic field is a region or space around a magnet or current carrying conductor in which its influence can be experienced. Magnetic field is a vector quantity. SI unit of the magnetic field is Tesla and the CGS unit of the magnetic field is Gauss. The relation between SI unit and CGS unit of magnetic field is given as $1{\text{T = 1}}{{\text{0}}^{\text{4}}}{\text{G}}$.
For finding the ratio of magnetic field at the centre of a current carrying coil of radius ${\text{R}}$ and at a distance of ${\text{3 R}}$ on its axis is given by using relation
 \[\dfrac{{{{\text{B}}_{{\text{centre}}}}}}{{{{\text{B}}_{{\text{axis}}}}}}{\text{ = }}{\left( {{\text{1 + }}\dfrac{{{x^2}}}{{{{\text{R}}^{\text{2}}}}}} \right)^{{\text{3/2}}}}\]
Substituting the value of ${\text{R}}$ and ${\text{x}}$ in the above formula, we get
\[
  \dfrac{{{{\text{B}}_{{\text{centre}}}}}}{{{{\text{B}}_{{\text{axis}}}}}}{\text{ = }}{\left( {{\text{1 + }}\dfrac{{{{{\text{(R)}}}^{\text{2}}}}}{{{{{\text{(3R)}}}^{\text{2}}}}}} \right)^{{\text{3/2}}}} \\
  \dfrac{{{{\text{B}}_{{\text{centre}}}}}}{{{{\text{B}}_{{\text{axis}}}}}}{\text{ = }}{\left( {{\text{1 + }}\dfrac{{{{\text{R}}^{\text{2}}}}}{{{\text{9}}{{\text{R}}^{\text{2}}}}}} \right)^{{\text{3/2}}}} \\
   \Rightarrow \dfrac{{{{\text{B}}_{{\text{centre}}}}}}{{{{\text{B}}_{{\text{axis}}}}}}{\text{ = }}{\left( {\dfrac{{{\text{10}}}}{1}} \right)^{{\text{3/2}}}} \\
  \therefore {\text{ }}\dfrac{{{{\text{B}}_{{\text{centre}}}}}}{{{{\text{B}}_{{\text{axis}}}}}}{\text{ = 10}}\sqrt {{\text{10}}} \\
 \]
Thus, ratio of magnetic field at the centre of a current carrying coil of radius ${\text{R}}$ and at a distance of ${\text{3 R}}$ on its axis is ${\text{10}}\sqrt {{\text{10}}} $.

Therefore, option (A) is the correct choice.

Note: When a charged particle enters in a region where both electric field ${\text{E}}$ and magnetic field B exist then force act on the charged particle is called Lorentz force. The formula of the Lorentz force is given by \[F = q[\mathop E\limits^ \to + \mathop v\limits^ \to \times \mathop B\limits^ \to ]\]. Velocity selector is an arrangement of crossed electric and magnetic fields such that when charge particles move in this region then the net force acting on it is zero.