Animated Solution for Physics - Magnetic Effects of Current: 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 +j^-direction. Which of the following statements regarding the magnetic field B is (are) true?
Select Answer:
* Multiple Correct
Visualized Solution
System Setup
Two infinite wires at x=±R.
Ring of radius R at z=3R.
Bring at Origin
Current in ring is clockwise from above.
By Right-Hand Rule, Bring at origin is along −k^.
Bring=0
Option (a): I1=I2
Bwires at origin =2πRμ0I1k^+2πRμ0I2(−k^)=0
Bnet=Bring+Bwires=Bring=0
Option (a) is True.
Option (b): I1>0,I2<0
Bwires at origin is along +k^.
Bring is along −k^.
They can cancel each other out, so Bnet can be zero.
Option (b) is True.
Option (c): I1<0,I2>0
Bwires at origin is along −k^.
Bring is also along −k^.
They add up, so Bnet cannot be zero.
Option (c) is False.
Option (d): B at C(0,0,3R)
For I1=I2, the z-components of B from the two wires perfectly cancel at C.
Bwires at C is purely along the x-axis.
The only z-component is due to the ring: Bring=−2Rμ0Ik^
Option (d) is True.
Final Conclusion
Correct Options: (a), (b), (d)
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The Sigma Insight: Biot-Savart Law
Solution Diagram
The problem presents a beautiful 3D arrangement of current-carrying elements: two infinitely long straight wires and a circular ring. Our goal is to analyze the magnetic field B at specific points in space under different current conditions. Let's break down the physics step-by-step.
Analyzing the Setup
We have two straight wires lying in the xy-plane at x=+R and x=−R. A positive current in these wires means it flows in the +j^ direction. Above them, a circular ring of radius R is suspended at a height of z=3R, carrying a current I in the clockwise direction when viewed from above.
The first crucial observation is the magnetic field produced by the ring at the origin (0,0,0). Using the Right-Hand Thumb Rule, if we curl our fingers in the clockwise direction, our thumb points downwards. Therefore, the magnetic field due to the ring at the origin, Bring, is directed along the −k^ direction. This field is a constant non-zero vector in our analysis.
Evaluating the Options at the Origin
Now, let's test the given options by analyzing the magnetic field produced by the two straight wires at the origin, Bwires.
Option (a): I1=I2
If both wires carry equal currents in the same direction, their magnetic fields at the origin will perfectly cancel each other out. Wire 1 (at x=+R) produces a field in the +k^ direction, while Wire 2 (at x=−R) produces a field in the −k^ direction.
Bwires=2πRμ0I1k^+2πRμ0I2(−k^)=0
However, the net magnetic field is the sum of the fields from the wires and the ring. Since $\mathbf{B}_{\text{ring}}
eq 0$, the net field B cannot be zero. Thus, Option (a) is correct.
Option (b): I1>0 and I2<0
Here, Wire 1 carries current in the +j^ direction, producing a field in the +k^ direction at the origin. Wire 2 carries current in the −j^ direction, which also produces a field in the +k^ direction at the origin.
Therefore, Bwires points in the +k^ direction. Since Bring points in the −k^ direction, it is entirely possible for these two opposing fields to have equal magnitudes and cancel each other out, resulting in a net zero field. Thus, Option (b) is correct.
Option (c): I1<0 and I2>0
In this scenario, the currents are reversed compared to option (b). Both wires now produce magnetic fields in the −k^ direction at the origin.
Since Bring is also in the −k^ direction, all three magnetic field vectors point downwards. They will add up, meaning the net magnetic field can never be zero. Thus, Option (c) is incorrect.
The Climax at the Center of the Ring
Option (d): I1=I2
We need to find the z-component of the net magnetic field at the center of the loop, C(0,0,3R).
Let's look at the symmetry of the setup. The two wires are placed symmetrically at x=+R and x=−R, and they carry equal currents. The magnetic field vectors they produce at point C will tilt symmetrically.
Specifically, the z-component of the field from Wire 1 will be exactly equal and opposite to the z-component of the field from Wire 2. They perfectly cancel each other out!
(Bwires)z=0
Therefore, the only contribution to the z-component of the magnetic field at C comes from the ring itself. The magnetic field at the center of a current-carrying ring is given by:
Bring=2Rμ0I(−k^)
So, the z-component is indeed −2Rμ0I. Thus, Option (d) is correct.
Final Conclusion
By carefully applying the Right-Hand Rule and leveraging the spatial symmetry of the system, we have successfully navigated through the options