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Question: The diagram shows two antiparallel currents that are equal in magnitude. AB is a line perpendicular ...

The diagram shows two antiparallel currents that are equal in magnitude. AB is a line perpendicular to both the currents and O is its center. Consider now the following statements
(i) The resultant magnetic field at O is non-zero.
(ii) The magnitude of the magnetic field along AB is constant.
(iii) The direction of the magnetic field is the same everywhere on AB.
Which of the following is correct?

(A) (i), (ii) and (iii)
(B) (i) and (ii)
(C) (i) and (iii)
(D) (ii) and (iii)

Explanation

Solution

To solve this question, we need to use the ampere’s circuital law in order to determine the magnitude of the electric field between the currents. For finding out the direction of the magnetic field, we can use the right hand rule.

Formula used: The formula which is used in solving this question is given by
B=μ0I2πrB = \dfrac{{{\mu _0}I}}{{2\pi r}}, here BB is the magnetic field produced by a straight current carrying conductor at a perpendicular distance of rr from it.

Complete step by step solution:
Let the distance between the two currents be dd and the magnitude of the current be II. Now, let us consider a point C on the line AB, at a distance of xx from the upper current, as shown in the figure below.

The magnetic field due to the upper current at C is given by
B1=μ0I2πr{B_1} = \dfrac{{{\mu _0}I}}{{2\pi r}} (1)
Also, the magnetic field due to the lower current at C is
B2=μ0I2π(dr){B_2} = \dfrac{{{\mu _0}I}}{{2\pi \left( {d - r} \right)}} (2)
By the right hand thumb rule, we get the direction of both B1{B_1} and B2{B_2}to be into the plane of the paper. Therefore the total magnetic field at C is
B=B1+B2B = {B_1} + {B_2}
From (1) and (2)
B=μ0I2πr+μ0I2π(dr)B = \dfrac{{{\mu _0}I}}{{2\pi r}} + \dfrac{{{\mu _0}I}}{{2\pi \left( {d - r} \right)}}
B=μ0I2π(1r+1(dr))B = \dfrac{{{\mu _0}I}}{{2\pi }}\left( {\dfrac{1}{r} + \dfrac{1}{{\left( {d - r} \right)}}} \right)
On simplifying, we get
B=μ0I2π(r+(dr)r(dr))B = \dfrac{{{\mu _0}I}}{{2\pi }}\left( {\dfrac{{r + \left( {d - r} \right)}}{{r\left( {d - r} \right)}}} \right)
B=μ0Id2πr(dr)B = \dfrac{{{\mu _0}Id}}{{2\pi r\left( {d - r} \right)}} (3)
So as we can see that the magnetic field between A and B is dependent on the distance rr. Therefore it is not constant along AB.
Thus the statement (ii) is incorrect.
Also, this magnetic field is into the plane of the paper, regardless of the distance rr. So, its direction is the same everywhere on AB.
So the statement (iii) is correct.
Now, as the point O lies at the centre of the line AB, we have r=d/2r = d/2 for this point.
Substituting r=d/2r = d/2 in (3) we get the magnetic field at O as
B=μ0Id2π(d/2)(dd/2)B = \dfrac{{{\mu _0}Id}}{{2\pi \left( {d/2} \right)\left( {d - d/2} \right)}}
On simplifying we get
B=2μ0IπdB = \dfrac{{2{\mu _0}I}}{{\pi d}}
So the magnetic field at O is clearly non-zero.
So the statement (i) is correct.
Thus, the correct statements are (i) and (iii).
Hence, the correct answer is option C.

Note:
We can attempt this question without even calculating the magnitude of the magnetic field. As the magnetic field produced by both the currents is in the same direction, so just with the help of the right hand thumb rule we can answer this question.