Q.A straight conducting wire PQ slides smoothly, without changing its orientation, over two long straight parallel conducting rails that lie in the plane of the paper and are separated by a perpendicular distance d. The plane of the rails carries a uniform magnetic field of magnitude B directed out of the paper. The sliding wire PQ is held at a fixed angle θ to the rails (it is not perpendicular to them) and is pushed along the rails with a constant velocity v directed along the length of the rails (perpendicular to their separation). The wire PQ has negligible resistance, and the only resistance in the closed circuit is R. Find the current in the circuit.

Concept

For a conductor moving in a magnetic field, the motional emf is $\varepsilon=\displaystyle\int(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ taken along the conductor; equivalently $\varepsilon=-\dfrac{d\phi}{dt}$, and the induced current is $I=\varepsilon/R$.

Setting up

Take $\mathbf{v}=v\,\hat{x}$ (along the rails) and $\mathbf{B}=B\,\hat{z}$ (out of the paper). Then

The wire PQ runs from one rail to the other; whatever its tilt θ, its two ends are separated in the $\hat{y}$ direction by exactly the rail spacing $d$. Writing $d\boldsymbol{\ell}$ for an element of PQ,

$\varepsilon=\int(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}=-vB\int dy=-vB\,d,$ …

Method: Computing Motional EMF When the Sliding Conductor Is Not Perpendicular to the Rails

Use this whenever a rod/wire slides along two parallel rails but is held at some angle $\theta$ to the rails, rather than straight across.

Steps

Step 1: Set up vectors for velocity and field

Write $\vec{v}$ along the direction of motion (along the rails) and $\vec{B}$ perpendicular to the plane, and compute the Lorentz-force-per-charge vector:

This vector always points along the rail-separation direction (perpendicular to $\vec{v}$), regardless of how the conductor itself is tilted.

Step 2: Integrate along the ACTUAL conductor, not the rail separation

taken along the wire. Because $(\vec{v}\times\vec{B})$ only has a component along the rail-separation direction, only the projection of the wire onto that direction contributes to the integral — any component of the wire running parallel to the rails contributes nothing (it's perpendicular to $\vec{v}\times\vec{B}$).

Step 3: Recognize the projection equals the fixed rail spacing

However the wire is tilted, both of its endpoints still sit on the two rails, so its projection onto the rail-separation direction is always exactly the rail spacing $d$ — the tilt angle $\theta$ never enters the final emf.

Step 4: Cross-check with the flux method …