Two long, straight wires carry equal currents in opposite directions as shown in the figure. The separation between the wires is 5.0 cm$5.0\text{ cm}$. The magnitude of the magnetic field at a point P$P$ midway between the wires is ______ $\mu\text{T}$. (Given : μ₀ = 4π × 10⁻⁷ T ⁻¹$\mu_{0} = 4\pi \times 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$, and each wire carries a current of 10 A$10\text{ A}$).

Related Formula

Core Logic

For a uniform magnetic field, the net force on an arbitrary shaped wire carrying a steady current depends only on its initial and final position. It is equivalent to the force on a straight wire connecting its ends.

The effective length $\vec{\ell}$ is a straight line joining the entry and exit points in the magnetic field. Length of equivalent straight wire, | | = 2R$|\vec{\ell}| = 2R$. Based on the current direction, it points in the +x$+x$ direction, so = 2R i$\vec{\ell} = 2R\hat{i}$.

Step 2: Cross Product Calculation

The magnetic field is given as B = B₀ k$\vec{B} = B_0 \hat{k}$ (from the visual diagram showing dot outwards along the z$z$-axis, though text incorrectly labeled it j$\hat{j}$, the intended field matches the standard coordinate system for such setups where force pushes up/down. Wait, following the PDF's solution logic explicitly:)

Solution specifies: Note: Direction of magnetic field is in +k$+k$ due to visual dot convention. So B = B k$\vec{B} = B \hat{k}$.

Since i × k = - j$\hat{i} \times \hat{k} = -\hat{j}$:

Pattern Recognition

Replace any semicircular current loop with its straight line displacement vector 2R$2R$. Then just take L × B$\vec{L} \times \vec{B}$. The visual dots clearly represent + k$+\hat{k}$.

Chapter Mix

Class 12 Physics: Moving Charges And Magnetism

Related Formula

Core Logic

For an electron, the charge is q = -e$q = -e$. The vector cross product generates the magnetic force. (Note: The PDF solution uses q = e$q = e$ implicitly for magnitude, but taking the full vector product is required. Let's trace it exactly).

F = e ( v × B)$$\vec{F} = e (\vec{v} \times \vec{B})$$ (Using q=e$q=e$ as per the PDF's sign convention for the variable e$e$, representing the base charge value)

Step 1: Expanding Cross Product

Step 2: Final Calculation

Substitute back into the force equation:

Chapter Mix

Class 12 Physics: Moving Charges And Magnetism