The current of 5 ~A$5 \mathrm{~A}$ flows in a square loop of sides 1 ~m$1 \mathrm{~m}$ is placed in air. The magnetic field at the centre of the loop is X√(2) × 10⁻⁷ T$\mathrm{X}\sqrt{2} \times 10^{-7} \mathrm{T}$. The value of X$\mathrm{X}$ is ________

Related Formula

Core Logic

The two loops are mutually perpendicular to each other. The magnetic field at the center due to each loop acts along their respective normal axes, meaning the two resulting magnetic field vectors are orthogonal (90^°$90^\circ$ apart).

Step 1: Calculate Individual Fields

Magnetic field at the center due to loop A:

Magnetic field at the center due to loop B:

Step 2: Resultant Magnetic Field

Since BA$B_A$ and BB$B_B$ are perpendicular to each other:

Pattern Recognition

When two identical physical sources generate orthogonal vector fields of magnitude B$B$, the resultant is simply √(2)B$\sqrt{2}B$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

Core Logic

Earth's horizontal magnetic field (BH$B_H$) points strictly from South to North. The conductor is aligned from South-East to North-West. Thus, the angle θ$\theta$ between the current flow direction (SE to NW) and Earth's magnetic field (S to N) is exactly 45^°$45^\circ$.

Step 1: Compute Force per Unit Length

BH = 3.5 × 10⁻⁵ ~T$B_H = 3.5 \times 10^{-5} \mathrm{~T}$ i = √(2) ~A$i = \sqrt{2} \mathrm{~A}$ θ = 45^°$\theta = 45^\circ$

Step 2: Extract Value

The required multiplier for 10⁻⁶$10^{-6}$ is 35.

Pattern Recognition

Geographical alignments often yield clean angles like 45^°$45^\circ$ or 90^°$90^\circ$. S-N field intersecting SE-NW current creates a 45^°$45^\circ$ angular intersection, perfectly canceling the √(2)$\sqrt{2}$ amp current.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Magnetism and Matter

Related Formula

Core Logic

The area vector A$\vec{A}$ is perpendicular to the plane of the loop. According to the right-hand rule, an anticlockwise current observed from the negative X-axis means the area vector points in the negative X-direction.

Step 1: Calculating Magnetic Moment

Area of the loop, A = 0.2 m × 0.1 m = 0.02 m²$A = 0.2 \text{ m} \times 0.1 \text{ m} = 0.02 \text{ m}^2$

Magnetic moment vector:

Step 2: Calculating Torque

The magnetic field is B = 2 × 10⁻³ j T$\vec{B} = 2 \times 10^{-3} \hat{j} \text{ T}$.

Torque on the loop:

Step 3: Final Direction

The - k$-\hat{k}$ direction corresponds to the negative Z-direction. Magnitude is 2 × 10⁻⁴ N-m$2 \times 10^{-4} \text{ N-m}$.

Pattern Recognition

Identify the plane of the loop to get the possible normal vectors. Since the loop is in the Y-Z plane, A$\vec{A}$ must be ± i$\pm\hat{i}$. 'Anticlockwise from -X' fixes it as - i$-\hat{i}$. Simply apply cross product - i × j = - k$-\hat{i} \times \hat{j} = -\hat{k}$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

Bₙₑₜ = √(B₁² + B₂²)$$B_{\text{net}} = \sqrt{B_1^2 + B_2^2}$$ (for mutually perpendicular planes)

Core Logic

Calculate the magnetic field generated by each coil at their common center. Since the planes of the coils are perpendicular, their axial magnetic field vectors are also mutually perpendicular.

Step 1: Calculate Individual Fields

Radius r = π cm = π × 10⁻² m$r = \pi \text{ cm} = \pi \times 10^{-2} \text{ m}$. N = 100$N = 100$ iP = 1 A$i_P = 1 \text{ A}$, iQ = 2 A$i_Q = 2 \text{ A}$.

For coil P:

For coil Q:

Step 2: Calculate Resultant Field

Step 3: Extract x

Given format is √(x) mT$\sqrt{x} \text{ mT}$, so x = 20$x = 20$.

Pattern Recognition

For concentric perpendicular coils of same radius and turns, Bₙₑₜ = B₀ √(i₁² + i₂²)$B_{\text{net}} = B_0 \sqrt{i_1^2 + i_2^2}$ where B₀$B_0$ is the base field for 1 A. This speeds up calculation considerably.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism