Q.Deduce expression for electromotive force induced across the ends of a conducting rod moving in an uniform magnetic field. OR Describe a transformer under the following headings:

The key idea here is the distinction between induced emf and induced current. Many students mix them up, especially when a problem mentions resistance. Let's clear that up first.

When a conductor moves in a magnetic field, the free electrons inside it experience a magnetic Lorentz force $q(\vec{v} \times \vec{B})$. This force pushes charges along the conductor, creating a potential difference — that potential difference is the motional emf. It is a direct consequence of the motion and the field, nothing else.

The resistance of the conductor only comes into the picture when you ask: "How much current flows as a result of this emf?" That's Ohm's law: $I = \mathcal{E}/R$. But the emf itself is independent of the resistance.

Now let's walk through the problem step by step.

1. Identify the source of emf. The arm PQ is the only part of the loop that is moving. The rest of the loop (PSRQ) is stationary and has negligible resistance. So the entire induced emf in the loop is generated across PQ.

2. Write the expression for motional emf. For a straight conductor of length $l$ moving with velocity $v$ perpendicular to a uniform magnetic field $B$, the motional emf is:

This is derived from the work done per unit charge by the magnetic force: $F_m = q v B$, so the electric field set up inside the conductor is $E = vB$, and over length $l$, the potential difference is $E l = B l v$.

3. Check each option against this formula.
   • (a) magnetic field $B$ — appears in $\mathcal{E} = B l v$. So emf does depend on it.
   • (b) velocity $v$ — appears directly. So emf does depend on it.
   • (d) length of PQ — that's $l$ in the formula. So emf does depend on it.
   • (c) resistance $r$ — does not appear in $\mathcal{E} = B l v$. So emf does not depend on it. …

Why motional emf depends on perpendicular motion

Motional emf arises when a conductor cuts through magnetic field lines. The physical picture: as the rod moves, the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ pushes charge carriers along the rod, creating a potential difference. This force—and hence the emf—depends critically on the component of velocity perpendicular to the magnetic field.

The motional emf in a straight rod is given by

For a uniform field and velocity, this simplifies to

where $\vec{l}$ is the length vector along the rod. The cross product $\vec{v} \times \vec{B}$ measures how much the velocity is perpendicular to the field. When $\vec{v}$ and $\vec{B}$ are parallel (or antiparallel), this cross product vanishes.

Step-by-step analysis

1. Identify the geometry

   The rod of length $l$ is parallel to $\vec{B}$, and it moves with velocity $\vec{v}$ along the direction of $\vec{B}$. So $\vec{v} \parallel \vec{B}$.

2. Compute the cross product

   Since $\vec{v}$ and $\vec{B}$ point in the same (or exactly opposite) direction, the angle $\theta$ between them is either $0°$ or $180°$. The magnitude of the cross product is

3. Evaluate the motional emf Substituting into the emf formula, …