Question

An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field \(B_0\), as shown in figure. The force on the loop due to magnetic field is

• A. \(ILB_0\cos\theta\)
• B. \(2ILB_0\cos\theta\)
• C. \(0\)
• D. \(ILB_0\sin\theta\)

Hint:

The net force on any closed current-carrying loop placed in a uniform magnetic field is zero, although the loop may experience torque.

Answer 1

The Correct Option is C

Step 1: Recall the force on a current-carrying wire segment.
The magnetic force on a straight wire of length $\vec{l}$ carrying current $I$ in a uniform field $\vec{B}$ is:
\[ \vec{F} = I(\vec{l} \times \vec{B}) \]

Step 2: Consider the concept of net displacement vector for the whole loop.
For the entire closed triangular loop, the sum of all the displacement vectors $\vec{l}_i$ (for each side, going around the loop) must equal zero, because the loop returns to the starting point.

Step 3: Write the total force on the closed loop.
Since the field $\vec{B}$ is uniform (does not depend on position), we can take it outside the sum:
\[ \vec{F}_\text{net} = I \left(\sum_i \vec{l}_i\right) \times \vec{B} \]

Step 4: Evaluate the sum of displacement vectors.
For a closed loop, the vector sum of all side lengths (with direction) is zero:
\[ \sum_i \vec{l}_i = \vec{0} \]

Step 5: Compute the net force.
\[ \vec{F}_\text{net} = I(\vec{0}) \times \vec{B} = \vec{0} \]

Step 6: State the answer.
No matter what the shape of the closed loop, the net force in a uniform magnetic field is always zero.
\[ \boxed{0} \]