Two infinitely long straight wires lie in the xy-plane along the lines x - - R- The wire located at x - -R carries a constant current I1 and the wire located at x -R carries a constant current I2- A circular loop of radius R is suspended with its centre at 0- 0- -3R and in a plane parallel to the xy-plane- This loop carries a constant current I in the clockwise direction as seen from above the loop- The current in the wire is taken to be positive if it is in the - direction- Which of the following statements regarding the magnetic field B is are true -

The correct options are A If I_1 = I_2 , then B cannot be equal to zero at the origin (0, 0, 0) C If I_1 > 0 and I_2 D If I_1 = I_2 ...

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Byju's Answer

Standard XII

Physics

Magnetic Field

Two infinitel...

Question

Two infinitely long straight wires lie in the $x y$ -plane along the lines $x = \pm R$ . The wire located at $x = + R$ carries a constant current $I_{1}$ and the wire located at $x = - R$ carries a constant current $I_{2}$ . A circular loop of radius $R$ is suspended with its centre at $(0, 0, \sqrt{3} R)$ and in a plane parallel to the $x y$ -plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+^j$ direction. Which of the following statements regarding the magnetic field $\to B$ is (are) true ?

I_{1} = I_{2}

, then

\to B

cannot be equal to zero at the origin

(0, 0, 0)

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I_{1} > 0

and

I_{2} < 0

, then

\to B

can be equal to zero at the origin

(0, 0, 0)

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I_{1} < 0

and

I_{2} > 0

, then

\to B

can be equal to zero at the origin

(0, 0, 0)

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I_{1} = I_{2}

, then the

z

-component of the magnetic field at the centre of the loop is

(- \frac{μ_{0} I}{2 R})

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Solution

The correct options are
A If $I_{1} = I_{2}$ , then $\to B$ cannot be equal to zero at the origin $(0, 0, 0)$
C If $I_{1} > 0$ and $I_{2} < 0$ , then $\to B$ can be equal to zero at the origin $(0, 0, 0)$
D If $I_{1} = I_{2}$ , then the $z$ -component of the magnetic field at the centre of the loop is $(- \frac{μ_{0} I}{2 R})$
(A) At origin, $\to B = 0$ due to two wires if $I_{1} = I_{2}$ (they cancel each other)
hence $({\to B}_{n e t})$ at origin is equal to $\to B$ due to ring, which is non-zero.
(B) If $I_{1} > 0$ and $I_{2} < 0, \to B$ at origin due to wires will be along $+^k$ direction and $\to B$ due to ring is along $-^k$ direction and hence $\to B$ can be zero at origin.
(C) If $I_{1} < 0$ and $I_{2} > 0, \to B$ at origin due to wires is along $-^k$ and also along $-^k$ due to ring, hence $\to B$ cannot be zero.
(D) (ref. image 2) At centre of ring, $\to B$ due to wires is along $x$ -axis,
hence $z$ -component is only because of ring which $\to B = \frac{μ_{0} I}{2 R} (-^K)$

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