Question

1 m long wire is folded in the form of a circular coil and 100 mA electric current is flowing in it, then magnetic field at a point 1 m away from its centre on its axis:
A. $\dfrac{{{{10}^{ - 5}}}}{{4\pi }}T$
B. $\dfrac{{{{10}^{ - 8}}}}{{2\pi }}T$
C. $\dfrac{{{{10}^{ - 5}}}}{{2\pi }}T$
D. $\dfrac{{{{10}^{ - 8}}}}{{4\pi }}T$

Answer 1

Hint: Charges at rest produce only electric fields while charges in motion produce both electric and magnetic fields. Current flowing through wire means that charge is in motion which produces the magnetic field and we will calculate the magnetic field produced by it at the required point.

Formula used:
$\eqalign{
  & r = \dfrac{l}{{2\pi }} \cr
  & B = \dfrac{{{\mu _0}i{r^2}}}{{2{{\left( {{r^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}} \cr} $

Complete step-by-step answer:
It is given that the wire of length l=1m is converted into circular form. The length of the circle will be equal to the length of the wire.
So the radius of the circle will be $r = \dfrac{l}{{2\pi }}$ where ‘r’ is the radius and ‘l’ is the length of the wire. So here the radius will be $r = \dfrac{l}{{2\pi }} = \dfrac{1}{{2\pi }}m$
Now we can assume this circular wire as a thin ring and we have a formula to find out the magnetic field on any point on the axis of the ring.
That formula is $B = \dfrac{{{\mu _0}i{r^2}}}{{2{{\left( {{r^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$ where ${\mu _0}$ is the permeability in free space and equal to $4\pi \times {10^{ - 7}}$ and ’i’ is the current and equal to $100mA = 100 \times {10^{ - 3}}A$ and ‘r’ is the radius which was already found and ‘x’ is the axial distance from the center of the circular wire which is given as one meter. By substituting all these values in the formula we get
$B = \dfrac{{{\mu _0}i{r^2}}}{{2{{\left( {{r^2} + {x^2}} \right)}^{\dfrac{3}{2}}}}}$
$\eqalign{
  & \Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times {{10}^{ - 3}} \times {{\left( {\dfrac{1}{{2\pi }}} \right)}^2}}}{{2{{\left( {{{\left( {\dfrac{1}{{2\pi }}} \right)}^2} + {1^2}} \right)}^{\dfrac{3}{2}}}}} \cr
  & But\ here \left( {{{\left( {\dfrac{1}{{2\pi }}} \right)}^2} + {1^2}} \right) \approx 1 \cr
  & \Rightarrow B = \dfrac{{{{10}^{ - 8}}}}{{2\pi }}T \cr} $

So, the correct answer is “Option B”.

Note: If we curl our right hand along the circular path of flow of current in circular wire then along the thumb we will get the direction of the magnetic field at a point on the axis of the ring. The magnetic field direction on any point on the axis of the ring will be along the axis because the other components will get cancelled as the ring is symmetrical.