Two infinitely long straight wires lie in the -plane along the lines The wire located at carries a constant current and the wire located at carries a constant current . A circular loop of radius is suspended with its centre at and in a plane paralle

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 $+ \hat{j}$ direction. Which of the following statements regarding the magnetic field $\vec{B}$ is (are) true? [2018]

Smart Achievers · Accepted Answer

(1, 2, 4)

(1) If $I_{1} = I_{2}$ , then the magnetic fields due to $x = + R$ and $I_{2}$ at origin $' O'$ will cancel out each other. But the magnetic field at $' O'$ due to the ring will be present. Therefore, $B$ cannot be zero at origin.

(2) If $I_{1} > 0$ and $I_{2} < 0$ , then the magnetic field due to both currents will be in the $+ Z$ direction and add up. The magnetic field due to current $I$ will be in the $- Z$ direction and if its magnitude is equal to the combined magnitudes of $I_{1}$ and $I_{2}$ , then $B$ can be zero at the origin.

(3) If $I_{1} < 0$ and $I_{2} > 0$ , then their net magnetic field at the origin will be in the $- Z$ direction and the magnetic field due to current $I$ at the origin will also be in the $- Z$ direction. Therefore, $\vec{B}$ at origin cannot be zero.

(4) If $I_{1} = I_{2}$ , then the resultant of the magnetic field $B_{R}$ at $P$ is along the $+ X$ direction. Therefore, the magnetic field at $P$ is only due to the current $I$ , which is in
the $- Z$ direction and is equal to $\vec{B} = \frac{μ_{0} I}{2 R} (- \hat{k})$

Solution

1 If $I_{1} = I_{2}$ , then $\vec{B}$ cannot be equal to zero at the origin $(0, 0, 0)$

2 If $I_{1} > 0$ and $I_{2} < 0$ , then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$

3 If $I_{1} < 0$ and $I_{2} > 0$ , then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$

4 If $I_{1} = I_{2}$ , then the $z$ -component of the magnetic field at the centre of the loop is $(- \frac{μ_{0} I}{2 R})$

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Solution

1 If I1=I2, then B→ cannot be equal to zero at the origin (0,0,0)

2 If I1>0 and I2<0, then B→ can be equal to zero at the origin (0,0,0)

3 If I1<0 and I2>0, then B→ can be equal to zero at the origin (0,0,0)

4 If I1=I2, then the z-component of the magnetic field at the centre of the loop is (-μ0I2R)

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1 If $I_{1} = I_{2}$ , then $\vec{B}$ cannot be equal to zero at the origin $(0, 0, 0)$

2 If $I_{1} > 0$ and $I_{2} < 0$ , then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$

3 If $I_{1} < 0$ and $I_{2} > 0$ , then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$

4 If $I_{1} = I_{2}$ , then the $z$ -component of the magnetic field at the centre of the loop is $(- \frac{μ_{0} I}{2 R})$