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Please solve this prob :-

Q1 A circular wire-loop of radius a carries a current i and is placed in perpendicular magnetic field B. Find the force of compression in the wire.

My answer coming to this is i (2 pi a) B. But, correct is iaB!!

Also, net force on a current carrying closed loop in a tranverse magnetic field is 0. Then, how come above a non-zero value is coming.

Let's break down the problem step by step to clarify the concepts involved and resolve the confusion regarding the force on the circular wire loop carrying a current in a magnetic field.

Understanding the Setup

We have a circular wire loop with a radius \( a \) carrying a current \( i \). This loop is placed in a magnetic field \( B \) that is perpendicular to the plane of the loop. The key here is to analyze how the magnetic field interacts with the current flowing through the wire.

Magnetic Force on a Current-Carrying Wire

The force \( \mathbf{F} \) on a segment of wire carrying a current \( i \) in a magnetic field \( \mathbf{B} \) can be calculated using the formula:

F = i (\mathbf{L} \times \mathbf{B})

Here, \( \mathbf{L} \) is the length vector of the wire segment, and the direction of the force is given by the right-hand rule. For a circular loop, we need to consider the entire loop rather than just a segment.

Calculating the Force on the Loop

For a circular loop, the total length \( L \) of the wire is the circumference of the circle, which is given by:

L = 2 \pi a

However, when we consider the force on the entire loop in a uniform magnetic field, we must remember that the forces acting on opposite sides of the loop will cancel each other out. This is because the magnetic force on one side of the loop is equal in magnitude but opposite in direction to the force on the directly opposite side.

Net Force on the Loop

As you correctly pointed out, the net force on a closed loop in a uniform magnetic field is indeed zero. This is due to the symmetry of the forces acting on the loop. Each segment of the wire experiences a force, but these forces balance out when considering the entire loop.

Force of Compression in the Wire

Now, regarding the force of compression in the wire, it is important to note that while the net force on the loop is zero, there can still be internal stresses within the wire due to the magnetic field. The magnetic field exerts a tension that tries to pull the wire outward, which can be interpreted as a compressive force acting along the wire itself.

To summarize, while your initial calculation of \( i (2 \pi a) B \) gives a value related to the forces acting on the wire, the correct interpretation in the context of a closed loop in a uniform magnetic field is that the net external force is zero. The internal stresses and forces of compression arise from the interactions of the magnetic field with the current, but they do not result in a net force acting on the loop as a whole.

Key Takeaways

• The net force on a closed current-carrying loop in a uniform magnetic field is zero.
• Forces on opposite sides of the loop cancel each other out.
• Internal stresses can exist within the wire due to the magnetic field, leading to a force of compression.

Understanding these concepts is crucial for grasping the behavior of current-carrying conductors in magnetic fields. If you have any further questions or need clarification on any specific point, feel free to ask!