Answer
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Hint: As a first step, you could make a rough diagram to understand the given situation. Then you could recall the standard expressions for both the given cases, that is, for the magnetic fields inside and outside respectively. You could then accordingly substitute and take the ratio to find the answer.
Formula used:
Magnetic field:
Inside the wire,
$B=\dfrac{{{\mu }_{0}}Ir}{2\pi {{R}^{2}}}$
Outside,
$B'=\dfrac{{{\mu }_{0}}I}{2\pi r}$
Complete Step by step solution:
In the question, we are given a long straight wire of radius a carrying the current I. we are also said that the current is uniformly distributed over the entire cross section. We are asked to find the ratio magnetic field at radial distances $\dfrac{a}{2}$ and 2a.
Let B be the magnetic field at radial distance $\dfrac{a}{2}$ which would be inside the wire and B’ be the magnetic field at radial distance 2a which would be outside of the wire.
We know that magnetic field at a distance $r\langle R$ from the centre of a wire of radius R is given by,
$B=\dfrac{{{\mu }_{0}}Ir}{2\pi {{R}^{2}}}$
Substituting the given values, we get,
$B=\dfrac{{{\mu }_{0}}I\left( \dfrac{a}{2} \right)}{2\pi {{a}^{2}}}$
$\therefore B=\dfrac{{{\mu }_{0}}I}{4\pi a}$ ………………………………………………….. (1)
Now for the second part let us recall the expression for magnetic field at distance $r\rangle R$ outside the wire of radius R will be,
$B'=\dfrac{{{\mu }_{0}}I}{2\pi r}$
Substituting the given values,
$B'=\dfrac{{{\mu }_{0}}I}{2\pi \left( 2a \right)}=\dfrac{{{\mu }_{0}}I}{4\pi a}$……………………………………………………. (2)
Now, we could take the ratio of (1) and (2) to get,
$\dfrac{B}{B'}=\dfrac{\left( \dfrac{{{\mu }_{0}}I}{4\pi a} \right)}{\left( \dfrac{{{\mu }_{0}}I}{4\pi a} \right)}=1$
Therefore, we found the ratio of magnetic fields at the given distances to be 1.
Hence, option C is found to be the correct answer.
Note:
You may have noticed that we have substituted the values in the respective standard equations for each case. You could easily derive them by using Ampere’s circuital law if you wish. Though by hearting formulae shouldn’t be promoted, if you could remember them, then it would save you time in competitive exams.
Formula used:
Magnetic field:
Inside the wire,
$B=\dfrac{{{\mu }_{0}}Ir}{2\pi {{R}^{2}}}$
Outside,
$B'=\dfrac{{{\mu }_{0}}I}{2\pi r}$
Complete Step by step solution:
In the question, we are given a long straight wire of radius a carrying the current I. we are also said that the current is uniformly distributed over the entire cross section. We are asked to find the ratio magnetic field at radial distances $\dfrac{a}{2}$ and 2a.
Let B be the magnetic field at radial distance $\dfrac{a}{2}$ which would be inside the wire and B’ be the magnetic field at radial distance 2a which would be outside of the wire.
We know that magnetic field at a distance $r\langle R$ from the centre of a wire of radius R is given by,
$B=\dfrac{{{\mu }_{0}}Ir}{2\pi {{R}^{2}}}$
Substituting the given values, we get,
$B=\dfrac{{{\mu }_{0}}I\left( \dfrac{a}{2} \right)}{2\pi {{a}^{2}}}$
$\therefore B=\dfrac{{{\mu }_{0}}I}{4\pi a}$ ………………………………………………….. (1)
Now for the second part let us recall the expression for magnetic field at distance $r\rangle R$ outside the wire of radius R will be,
$B'=\dfrac{{{\mu }_{0}}I}{2\pi r}$
Substituting the given values,
$B'=\dfrac{{{\mu }_{0}}I}{2\pi \left( 2a \right)}=\dfrac{{{\mu }_{0}}I}{4\pi a}$……………………………………………………. (2)
Now, we could take the ratio of (1) and (2) to get,
$\dfrac{B}{B'}=\dfrac{\left( \dfrac{{{\mu }_{0}}I}{4\pi a} \right)}{\left( \dfrac{{{\mu }_{0}}I}{4\pi a} \right)}=1$
Therefore, we found the ratio of magnetic fields at the given distances to be 1.
Hence, option C is found to be the correct answer.
Note:
You may have noticed that we have substituted the values in the respective standard equations for each case. You could easily derive them by using Ampere’s circuital law if you wish. Though by hearting formulae shouldn’t be promoted, if you could remember them, then it would save you time in competitive exams.
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