Let's solve the problem **step by step**: --- ### **Problem 2 Analysis** A **sheet of current** (with current per unit length η, A/m) is coming out of the page. The figure shows magnetic field vectors above and below the sheet, with equal magnitudes but opposite directions. --- ## **(a) Find \(\Delta B_x\) in terms of η** ### **Step 1: Apply Ampère's Law** Ampère's law for a path crossing the sheet (take a rectangular loop perpendicular to the sheet): \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \] - The current sheet has current per unit length η (A/m). - For a path crossing the sheet, let the length parallel to the sheet be \( L \). - The total current enclosed: \( I_{\text{enc}} = \eta L \). ### **Step 2: Contribution to the Integral** - **Above the sheet**: \( B_{\text{above}} \) - **Below the sheet**: \( B_{\text{below}} \) - The field is parallel to the path on both sides, but in opposite directions. Total line integral: \[ \oint \vec{B} \cdot d\vec{l} = (B_{\text{above}} - B_{\text{below}})L \] Set equal to \( \mu_0 \eta L \): \[ (B_{\text{above}} - B_{\text{below}})L = \mu_0 \eta L \] \[ B_{\text{above}} - B_{\text{below}} = \mu_0 \eta \] ### **Step 3: Express \(\Delta B_x\)** \[ \boxed{\Delta B_x = \mu_0 \eta} \] Where \( \Delta B_x \) is the difference in the horizontal (x) component of the magnetic field between the top and bottom of the sheet. --- ## **(b) Pressure or Tension on the Sheet** ### **Symmetric Case Reasoning** - In the symmetric case: \( B_{\text{above}} = -B_{\text{below}} \) - The magnetic field **exerts a force on the current sheet**. - Magnetic pressure is given by: \( P = \frac{B^2}{2\mu_0} \) - The field above pushes down; the field below pushes up. - **Net effect**: The sheet experiences a tension, trying to pull it apart (like the sides of a current loop repel). ### **What if the Sheet Has No Structural Strength?** - The sheet will **expand/disperse outward** due to the mutual repulsion of current elements (each bit of current repels its neighbor). - If it's not held together, it will **fly apart**, like a wire "exploding" from its own current's magnetic pressure. --- ## **Final Answers** ### **(a)** \[ \boxed{\Delta B_x = \mu_0 \eta} \] ### **(b)** - The sheet experiences an **outward tension** due to magnetic pressure. - If the sheet has no structural strength, it will **fly apart/disperse** due to this tension.

Let's carefully analyze and solve each part of the problem with detailed steps and explanations. --- ## **Given Data & Concepts** - The sheet carries a **current per unit length** \(\eta\) (units: A/m), coming out of the page. - The magnetic field vectors are shown above and below the sheet, with equal magnitudes but opposite directions. - The magnetic field difference across the sheet is \(\Delta B_x = B_{\text{above}} - B_{\text{below}}\). - We can add an arbitrary uniform magnetic field \( \mathbf{B}_0 \), but only the *difference* \(\Delta B_x\) is physically meaningful for the force calculations. --- ## **Part (a): Find \(\Delta B_x\) in terms of \(\eta\)** ### **Step 1: Recall Ampère’s Law** Ampère’s Law in differential form relates magnetic field to current: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \] - For a sheet of current with **current per unit length** \(\eta\), the **enclosed current** when crossing the sheet over length \(L\) is: \[ I_{\text{enc}} = \eta \times L \] - The magnetic field \(\vec{B}\) is tangential and approximately uniform just above and below the sheet. ### **Step 2: Choose an Amperian Loop** - The loop is a rectangle crossing the sheet perpendicularly, with sides: - Length \(L\) **parallel** to the sheet (say along the x-direction). - Short segments of length \(h\) **perpendicular** to the sheet, above and below. - The magnetic field is constant in the regions just above and below. - The contributions from the sides parallel to the sheet are: \[ B_{\text{above}} \times L \quad \text{and} \quad - B_{\text{below}} \times L \] - The contributions from the vertical sides (perpendicular to the sheet) cancel out because the field is parallel to the sides and doesn't do work in the perpendicular direction. ### **Step 3: Write the Ampère Loop Integral** \[ \oint \vec{B} \cdot d\vec{l} = B_{\text{above}} \times L - B_{\text{below}} \times L = \mu_0 \times (\eta L) \] Dividing both sides by \(L\): \[ B_{\text{above}} - B_{\text{below}} = \mu_0 \eta \] ### **Final result for part (a):** \[ \boxed{ \boxed{ \Delta B_x = B_{\text{above}} - B_{\text{below}} = \mu_0 \eta } } \] --- ## **Part (b): Magnetic pressure and the behavior of the sheet** ### **Step 1: Symmetric magnetic field configuration** - The problem states "the symmetric case shown"—meaning: \[ B_{\text{above}} = - B_{\text{below}} = B \] - From part (a): \[ B_{\text{above}} - B_{\text{below}} = \mu_0 \eta \] - Substituting \( B_{\text{below}} = - B_{\text{above}} = - B \): \[ B - (- B) = 2 B = \mu_0 \eta \] - Therefore: \[ B = \frac{\mu_0 \eta}{2} \] ### **Step 2: Magnetic pressure** - The **magnetic pressure** exerted by a magnetic field \(B\) is: \[ P = \frac{B^2}{2 \mu_0} \] - Substituting \( B = \frac{\mu_0 \eta}{2} \): \[ P = \frac{1}{2 \mu_0} \left( \frac{\mu_0 \eta}{2} \right)^2 = \frac{1}{2 \mu_0} \times \frac{\mu_0^2 \eta^2}{4} = \frac{\mu_0 \eta^2}{8} \] ### **Step 3: Force per unit area and tension** - The magnetic pressure acts perpendicular to the sheet, trying to **push the sheet outward**. - The net force (per unit area) tending to **expand or repel the sheet** is: \[ F/A = P = \frac{\mu_0 \eta^2}{8} \] ### **Step 4: If the sheet has no structural strength** - Without any mechanical support, this outward magnetic pressure will cause the sheet to **expand or disperse**. - Physically, the magnetic repulsion tends to **push the current elements apart**, leading the sheet to **fly apart**, similar to how a current-carrying wire or plasma might behave. --- ## **Final Answers** ### **(a) Difference in magnetic field across the sheet:** \[ \boxed{ \Delta B_x = \mu_0 \eta } \] ### **(b) Magnetic pressure/tension on the sheet:** - The magnetic pressure is: \[ P = \frac{\mu_0 \eta^2}{8} \] - If the sheet **lacks any structural strength**, this magnetic pressure will cause it to **expand or disintegrate** due to mutual repulsion. --- **Summary:** | Part | Result | Explanation | |---------|---------|--------------| | (a) | \(\Delta B_x = \mu_0 \eta\) | Derived from Ampère’s law considering the current sheet. | | (b) | Magnetic pressure \(P = \frac{\mu_0 \eta^2}{8}\) | Based on the magnetic field in the symmetric case; causes outward force on the sheet. Without structural support, the sheet will disperse. | --- Let me know if you'd like further clarification or additional details!