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# A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle $\theta$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is:A) Proportional to$2\left( {180^\circ - \theta } \right)$.B) Inversely proportional to r.C) Zero, only if$\theta = 180^\circ$.D) Zero for all values of$\theta$.

Last updated date: 25th Feb 2024
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Hint: A magnetic field is defined as the magnetic influence of a moving charge particle or a magnetized material. A conductor is defined as a substance which can conduct electricity with the help free electrons are called conductors.

Formula Used:
The formula of the magnetic field due to an arc is given by,
$B = \left( {\dfrac{{{\mu _o} \cdot I}}{{2R}}} \right) \times \left( {\dfrac{\theta }{{2\pi }}} \right)$
Where the magnetic field is the B radius of the circular loop is R current through the conductor is I and the angle of the sector is$\theta$.

It is given in the problem that a battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R and one of the arcs AB of the ring subtends an angle $\theta$ at the centre and we need to find the value of the magnetic induction at the centre due to the current in the ring.
The formula of the magnetic field due to an arc is given by,
$B = \left( {\dfrac{{{\mu _o} \cdot I}}{{2R}}} \right) \times \left( {\dfrac{\theta }{{2\pi }}} \right)$
Where magnetic field is B radius of the circular loop is R the current through the conductor is I and the angle of the sector is$\theta$.
First of all let us calculate the value of magnetic field in the arc ABC.
$\Rightarrow {B_1} = \left( {\dfrac{{{\mu _o} \cdot {I_1}}}{{2R}}} \right) \times \left( {\dfrac{\theta }{{2\pi }}} \right)$………eq. (1)
And the direction is upwards.
Now let us calculate the value of the magnetic field at the centre due to arc ADC.
$\Rightarrow {B_2} = \left( {\dfrac{{{\mu _o} \cdot {I_2}}}{{2R}}} \right) \times \left( {\dfrac{{2\pi - \theta }}{{2\pi }}} \right)$………eq. (2)
And the direction is downwards.
The net magnetic field at the centre of the ring is equal to,
$\Rightarrow B = {B_1} - {B_2}$
Replacing value of the magnetic field from equation (1) and equation (2).
$\Rightarrow B = {B_1} - {B_2}$
$\Rightarrow B = \left( {\dfrac{{{\mu _o} \cdot {I_1}}}{{2R}}} \right) \times \left( {\dfrac{\theta }{{2\pi }}} \right) - \left( {\dfrac{{{\mu _o} \cdot {I_2}}}{{2R}}} \right) \times \left( {\dfrac{{2\pi - \theta }}{{2\pi }}} \right)$
$\Rightarrow B = \left( {\dfrac{{{\mu _o} \cdot {I_1}}}{{2R}}} \right) \cdot \left( {\dfrac{\theta }{{2\pi }}} \right) - \left( {\dfrac{{{\mu _o} \cdot {I_2}}}{{2R}}} \right) \cdot \left( {\dfrac{{2\pi - \theta }}{{2\pi }}} \right)$………eq. (3)
Since according to ohm’s law,
$\Rightarrow {I_1} = \dfrac{V}{{{R_1}}}$
$\Rightarrow {I_1} = \dfrac{V}{{\left( {\rho \dfrac{{{l_1}}}{A}} \right)}}$
$\Rightarrow {I_1} = \dfrac{V}{{\left( {\rho \cdot r \cdot \theta } \right)}}$………eq. (4)
Similarly,
$\Rightarrow {I_2} = \dfrac{V}{{\left( {\rho \dfrac{{{l_2}}}{A}} \right)}}$
$\Rightarrow {I_2} = \dfrac{{VA}}{{\left( {\rho \cdot r \cdot \theta } \right)}}$………eq. (5)
Replacing the value of ${I_1}$ and ${I_2}$ from the equation (4) and equation (5) into equation (3).
$\Rightarrow B = \left( {\dfrac{{{\mu _o} \cdot {I_1}}}{{2R}}} \right) \cdot \left( {\dfrac{\theta }{{2\pi }}} \right) - \left( {\dfrac{{{\mu _o} \cdot {I_2}}}{{2R}}} \right) \cdot \left( {\dfrac{{2\pi - \theta }}{{2\pi }}} \right)$
$\Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }}\left[ {\dfrac{{VA}}{{\rho r\theta }} \times \dfrac{\theta }{r} - \dfrac{{VA}}{{\rho r\left( {2\pi - \theta } \right)}} \times \dfrac{{\left( {2\pi - \theta } \right)}}{r}} \right]$
$\Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }}\left[ {\dfrac{{VA}}{{\rho r}} \times \dfrac{1}{r} - \dfrac{{VA}}{{\rho r}} \times \dfrac{1}{r}} \right]$
$\Rightarrow B = 0$.
The magnetic field at the centre of the ring is equal to$B = 0$.
The correct answer for this problem is option D.

Note: According to Ampere’s law any conductor which is carrying current creates a magnetic field around the conductor. If a coil is containing a current then there will exist a magnetic field around the coil due to the passing current. The direction of the magnetic field can be taken out by using the right hand thumb rule.