Question

The equilateral triangular frame has current 2A. The side of frame is \(4\sqrt{3}\) cm. Magnetic field at center C is:

• A. \(30\sqrt{3} \, \mu T\)
• B. \(10\sqrt{3} \, \mu T\)
• C. \(3\sqrt{10} \, \mu T\)
• D. \(10\sqrt{10} \, \mu T\)

Hint:

For any regular n-sided polygon loop, you can find the field at the center by calculating the field from one side and multiplying by n.
Memorizing the geometry of common shapes like the equilateral triangle (distance from centroid to side, angles) can speed up problem-solving.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the total magnetic field at the center (centroid) of an equilateral triangular loop carrying a current. The total field is the vector sum of the fields produced by the three straight wire segments of the triangle.
Step 2: Key Formula or Approach:
The magnetic field \(B\) at a perpendicular distance \(d\) from a finite straight wire carrying current \(I\) is given by:
\[ B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2) \] where \(\theta_1\) and \(\theta_2\) are the angles subtended by the ends of the wire at the point.
Step 3: Detailed Explanation:
Part A: Geometric Parameters
- Side length of the equilateral triangle, \(a = 4\sqrt{3}\) cm = \(4\sqrt{3} \times 10^{-2}\) m.
- The center C is the centroid. The perpendicular distance \(d\) from the centroid to any side is given by \(d = \frac{a}{2\sqrt{3}}\).
\[ d = \frac{4\sqrt{3} \times 10^{-2}}{2\sqrt{3}} = 2 \times 10^{-2} \text{ m} \] - For an equilateral triangle, the angles subtended by the ends of any side at the centroid are \(\theta_1 = 60^\circ\) and \(\theta_2 = 60^\circ\).
Part B: Magnetic Field from One Side
Using the formula for a finite wire with \(I = 2\) A:
\[ B_{side} = \frac{\mu_0 I}{4\pi d}(\sin 60^\circ + \sin 60^\circ) = \frac{\mu_0 I}{4\pi d}(2 \sin 60^\circ) \] \[ B_{side} = \frac{(4\pi \times 10^{-7}) \times 2}{4\pi \times (2 \times 10^{-2})} \left(2 \times \frac{\sqrt{3}}{2}\right) \] \[ B_{side} = \frac{2 \times 10^{-7}}{2 \times 10^{-2}} \times \sqrt{3} = 10^{-5} \sqrt{3} \text{ T} \] Part C: Total Magnetic Field
By the right-hand rule, the magnetic field from each of the three sides points in the same direction at the center (e.g., into the page). Therefore, the net magnetic field is the sum of the magnitudes of the fields from the three sides.
\[ B_{net} = 3 \times B_{side} = 3 \times (10^{-5} \sqrt{3}) \text{ T} = 3\sqrt{3} \times 10^{-5} \text{ T} \] To express this in microteslas (\(\mu T\)), we multiply by \(10^6\).
\[ B_{net} = (3\sqrt{3} \times 10^{-5}) \times 10^6 \, \mu T = 30\sqrt{3} \, \mu T \] Step 4: Final Answer:
The magnetic field at the center C is \(30\sqrt{3} \, \mu T\).