Uniform magnetic fields of different strengths (B₁$(\mathrm{B}_1$ and B₂)$\mathrm{B}_2)$ , both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m$m$ and charge q$q$ , at the interface at an instant, moves into the region 2 with velocity v$v$ and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?

(Consider the velocity of the particle to be normal to the magnetic field and B₂ > B₁$\mathrm{B}_2 > \mathrm{B}_1$ )

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