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

To find the magnetic field at the center of an equilateral triangular frame carrying a current, we can use the formula for the magnetic field at the center of a polygon due to current: 

The magnetic field \( B \) at the center of an equilateral triangle with side length \( a \) and carrying current \( I \) is given by:

\(B = \frac{\sqrt{3} \mu_0 I}{2 \pi a}\)

where:

• \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A\) is the permeability of free space,
• \( I = 2 \, A \) (given in the problem),
• \( a = 4\sqrt{3} \, cm = 4\sqrt{3} \times 10^{-2} \, m \)

Substitute the known values into the formula:

\(B = \frac{\sqrt{3} \times 4\pi \times 10^{-7} \times 2}{2 \pi \times 4\sqrt{3} \times 10^{-2}}\)

Simplify the expression:

\(B = \frac{8\sqrt{3} \times 10^{-7}}{8\sqrt{3} \times 10^{-2}}\)

\(B = 10^{-5} \, T \, = \, 10 \, \mu T\)

But since there are three sides contributing to the magnetic field at the center, multiply the result by 3:

\(B_{\text{total}} = 3 \times 10 \, \mu T = 30 \, \mu T\)

Finally, the magnetic field at the center of the triangular frame is:

\(B_{\text{center}} = 30\sqrt{3} \, \mu T\)

Thus, the correct answer is \(30\sqrt{3} \, \mu T\).