
The parts of two concentric circular arcs joined by two radial lines and carries current $i$. The arcs subtend an angle θ at the center of the circle. The magnetic field at the centre O, is
(A) $\dfrac{{{\mu _0}i(b - a)\theta }}{{4\pi ab}}$
(B) \[\dfrac{{{\mu _0}i(b - a)}}{{\pi - \theta }}\]
(C) $\dfrac{{{\mu _0}i\left( {b - a} \right)\theta }}{{\pi ab}}$
(D) $\dfrac{{{\mu _0}i(a - b)}}{{2\pi ab}}$
Answer
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Hint: Use the formula ${B_{arc}}(center) = \dfrac{{{\mu _0}i}}{{2r}}\left( {\dfrac{\theta }{{2\pi }}} \right)$ to get the magnetic field at the centre of the circular arc. Find the magnetic field generated due to the two circular arcs separately and then add them as the resultant would be the sum of the two.
Complete step-by-step answer:
The magnetic field at the centre of a circular with radius r is given by :
${B_{arc}}\left( {center} \right) = \dfrac{{{\mu _0}i}}{{2r}}\left( {\dfrac{\theta }{{2\pi }}} \right)$
Where ${\mu _0}$ is the permeability constant
$i$ is the current draw
$\theta $ is the angle subtended by the arcs at the center
Let us assume that “a” and “b” are the radial distances from the center to the periphery of the loops.
Now, the given condition says our system has two closed loops. The two loops have their own magnetic fields. we can only know the magnitude of the magnetic field, but we can’t know the signs of the magnetic field as the direction of the fields are not given. At this point, we can see one thing that we have a “minus” sign between the radial distances in the numerator of the options given. This means that there must be a difference in the directions of the fields.
$
\Delta {B_{arc}} = {B_{arc}}(a) - {B_{arc}}(b) \\
= \dfrac{{{\mu _0}i\theta }}{{4\pi a}} - \dfrac{{{\mu _0}i\theta }}{{4\pi b}} \\
= \dfrac{{{\mu _0}i\theta }}{{4\pi }}\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right) \\
= \dfrac{{{\mu _0}i(b - a)\theta }}{{4\pi ab}} \\
$
Hence, option (A) is correct .
Note: We don’t know actually what is the direction of current in the loops. The resultant magnetic field is the sum of the magnetic fields (considering the signs – clockwise or counter clockwise). We just observed the options and then only we came to a conclusion that there may be one of the loops which has clockwise current flowing in it as an anticlockwise direction is considered to be positive in assumptions. And then add those fields with their respective signs.
Complete step-by-step answer:
The magnetic field at the centre of a circular with radius r is given by :
${B_{arc}}\left( {center} \right) = \dfrac{{{\mu _0}i}}{{2r}}\left( {\dfrac{\theta }{{2\pi }}} \right)$
Where ${\mu _0}$ is the permeability constant
$i$ is the current draw
$\theta $ is the angle subtended by the arcs at the center
Let us assume that “a” and “b” are the radial distances from the center to the periphery of the loops.
Now, the given condition says our system has two closed loops. The two loops have their own magnetic fields. we can only know the magnitude of the magnetic field, but we can’t know the signs of the magnetic field as the direction of the fields are not given. At this point, we can see one thing that we have a “minus” sign between the radial distances in the numerator of the options given. This means that there must be a difference in the directions of the fields.
$
\Delta {B_{arc}} = {B_{arc}}(a) - {B_{arc}}(b) \\
= \dfrac{{{\mu _0}i\theta }}{{4\pi a}} - \dfrac{{{\mu _0}i\theta }}{{4\pi b}} \\
= \dfrac{{{\mu _0}i\theta }}{{4\pi }}\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right) \\
= \dfrac{{{\mu _0}i(b - a)\theta }}{{4\pi ab}} \\
$
Hence, option (A) is correct .
Note: We don’t know actually what is the direction of current in the loops. The resultant magnetic field is the sum of the magnetic fields (considering the signs – clockwise or counter clockwise). We just observed the options and then only we came to a conclusion that there may be one of the loops which has clockwise current flowing in it as an anticlockwise direction is considered to be positive in assumptions. And then add those fields with their respective signs.
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