Question

The current of 5 A flows in a square loop of sides 1 m placed in air. The magnetic field at the center of the loop is \(X\sqrt{2} \times 10^{-7} \, T\). The value of X = _________.

Answer 1

Correct Answer: 40

This task involves calculating the magnetic field at the center of a square loop carrying current and then determining a constant \( X \) from the derived magnetic field expression.

Concept Used:

The magnetic field \( B \) generated by a finite straight wire segment with current \( I \) at a perpendicular distance \( d \) is defined by the Biot-Savart law as:

\[B = \frac{\mu_0 I}{4\pi d} (\sin\theta_1 + \sin\theta_2)\]

Here, \( \theta_1 \) and \( \theta_2 \) represent the angles subtended by the wire segment's endpoints at the point of interest. For a square loop, the total magnetic field at its center is the vector sum of the fields from its four sides. Due to symmetry, the magnetic field from each side has the same magnitude and direction at the center.

\[B_{\text{center}} = 4 \times B_{\text{one side}}\]

Step-by-Step Solution:

Step 1: Identify Given Parameters.

Current in the loop: \( I = 5 \, \text{A} \).

Side length of the square loop: \( L = 1 \, \text{m} \).

Permeability of free space (for air): \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \).

Step 2: Determine Geometric Parameters for One Side.

The perpendicular distance \( d \) from the square's center to the midpoint of any side equals half the side length.

\[d = \frac{L}{2} = \frac{1}{2} = 0.5 \, \text{m}\]

The angles \( \theta_1 \) and \( \theta_2 \) are equal since the lines from the center to the vertices of a side form \( 45^\circ \) angles with the perpendicular distance due to the square's geometry.

\[\theta_1 = \theta_2 = 45^\circ\]

Step 3: Calculate Magnetic Field from One Side.

Using the formula for a finite wire:

\[B_{\text{one side}} = \frac{\mu_0 I}{4\pi d} (\sin 45^\circ + \sin 45^\circ)\]

Substituting values:

\[B_{\text{one side}} = \frac{(4\pi \times 10^{-7}) \times 5}{4\pi \times 0.5} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)\]\[B_{\text{one side}} = \frac{5 \times 10^{-7}}{0.5} \left(\frac{2}{\sqrt{2}}\right)\]\[B_{\text{one side}} = 10 \times 10^{-7} \times \sqrt{2} = \sqrt{2} \times 10^{-6} \, \text{T}\]

Step 4: Calculate Total Magnetic Field at the Center.

The total magnetic field is four times the field from a single side.

\[B_{\text{center}} = 4 \times B_{\text{one side}}\]\[B_{\text{center}} = 4 \times (\sqrt{2} \times 10^{-6}) = 4\sqrt{2} \times 10^{-6} \, \text{T}\]

Expressed in the given format:

\[B_{\text{center}} = 40\sqrt{2} \times 10^{-7} \, \text{T}\]

Step 5: Determine X by Comparing Fields.

The problem provides the magnetic field at the center as \( X\sqrt{2} \times 10^{-7} \, \text{T} \).

\[X\sqrt{2} \times 10^{-7} = 40\sqrt{2} \times 10^{-7}\]

Equating both sides yields:

\[X = 40\]

The value of X is 40.