Question

As shown in the figure, a uniform straight wire of length $30\sqrt{3}$ cm is bent in the form of an equilateral triangle ABC. A uniform magnetic field 2T is applied parallel to the side BC. If the current through the wire is 2A, the magnitude of the force on the side AC is

• A. $\frac{2}{\sqrt{3}}$ N
• B. $0.2\sqrt{3}$ N
• C. $1.2$ N
• D. $0.6$ N

Hint:

The magnetic force on a straight current-carrying wire segment is $\vec{F} = I(\vec{l} \times \vec{B})$. The magnitude $F = IlB\sin\theta$ is the key formula. In a closed loop in a uniform field, the net force is zero, but the force on an individual segment can be non-zero.

Answer 1

The Correct Option is D

Step 1: Determine the length of the side of the equilateral triangle.
The total length of the wire is $L_{total} = 30\sqrt{3}$ cm. The triangle has 3 equal sides, so the length of each side ($l$) is: \[ l = \frac{L_{total}}{3} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ cm}. \] Convert to SI units (meters): $l = 10\sqrt{3} \times 10^{-2} \text{ m} = 0.1\sqrt{3} \text{ m}$.

Step 2: State the formula for the magnetic force on a current-carrying wire.
The magnitude of the magnetic force $F$ on a segment of wire of length $l$ carrying a current $I$ in a uniform magnetic field $B$ is given by: \[ F = I l B \sin\theta, \] where $\theta$ is the angle between the current direction (along the wire) and the magnetic field $\vec{B}$.

Step 3: Determine the angle $\theta$ for the side AC.
The magnetic field $\vec{B}$ is applied parallel to the side BC. The angle between side AC and side BC in an equilateral triangle is $60^\circ$ (or $\pi/3$ rad). The current flows along the sides, and the magnetic field is parallel to BC. The angle $\theta$ between the current direction on side AC and the magnetic field (parallel to BC) is $60^\circ$.

Step 4: Substitute the given values and calculate the force.
Given values: $I=2$ A, $l=0.1\sqrt{3}$ m, $B=2$ T, and $\theta=60^\circ$. \[ F_{AC} = I l B \sin(60^\circ). \] \[ F_{AC} = (2 \text{ A}) (0.1\sqrt{3} \text{ m}) (2 \text{ T}) \left(\frac{\sqrt{3}}{2}\right). \] \[ F_{AC} = 2 \times 0.1 \times 2 \times \frac{3}{2} = 0.2 \times 3 = 0.6 \text{ N}. \] \[ \boxed{F_{AC} = 0.6 \text{ N}}. \]