Question

If a conducting rod of length $100\text{ cm}$ rotates about one of its ends with a constant frequency of $14\text{ revolutions per second}$ in a plane perpendicular to a uniform magnetic field of $2\text{ T}$, then the induced emf between the two ends of the rod is:

• A. $144\text{ V}$
• B. $88\text{ V}$
• C. $122\text{ V}$
• D. $230\text{ V}$

Hint:

For a rod rotating about one end in a uniform magnetic field, \[ \varepsilon=\frac{1}{2}B\omega L^2 \] or equivalently, \[ \varepsilon=\pi BfL^2 \] Memorize both forms. In MCQs, the second form is often quicker because frequency $f$ is usually given directly instead of angular velocity $\omega$.

Answer 1

The Correct Option is B

Step 1: Recall the rotating-rod emf.
A rod rotating about one end in a perpendicular field $B$ sweeps area and develops $\varepsilon = \tfrac12 B\omega L^2$, where $\omega$ is the angular speed and $L$ the rod length.
Step 2: Convert quantities.
$L = 100\,cm = 1\,m$, frequency $f = 14\,rev\,s^{-1}$, $B = 2\,T$.
Step 3: Find the angular velocity.
$\omega = 2\pi f = 2\pi(14) = 28\pi\,rad\,s^{-1}$.
Step 4: Substitute into the emf formula.
$\varepsilon = \tfrac12 (2)(28\pi)(1)^2 = 28\pi\,V$.
Step 5: Evaluate numerically.
Using $\pi \approx \tfrac{22}{7}$, $\varepsilon = 28\times \tfrac{22}{7} = 4\times 22 = 88\,V$.
Step 6: Conclude.
The induced emf between the ends of the rod is $88\,V$, which is option (2).
\[ \boxed{\varepsilon = 88\,V} \]