Question

A rod of mass $m$ and length $\ell$ is released from the position shown with its upper end hinged in a uniform horizontal magnetic field $B$. Find the maximum induced emf in the rod: 

• A. $B\ell\sqrt{\dfrac{3g\ell}{8}}$
• B. $B\ell\sqrt{\dfrac{g\ell}{8}}$
• C. $B\ell\sqrt{\dfrac{7g\ell}{8}}$
• D. $B\ell\sqrt{\dfrac{5g\ell}{8}}$

Hint:

For rotating conductors in a magnetic field, first find the maximum angular speed using energy conservation, then apply $\mathcal{E}_{\max}=\dfrac{1}{2}B\ell^2\omega$.

Answer 1

The Correct Option is A

To find the maximum induced emf in the rod, we can employ the concept of electromagnetic induction and rotational motion. The emf induced in a rotating rod in a magnetic field is given by:

\(\text{emf} = \frac{1}{2} B \omega \ell^2 \sin{\theta}\)

Where:

• \(B\) is the magnetic field.
• \(\omega\) is the angular speed.
• \(\ell\) is the length of the rod.
• \(\theta\) is the angle between the rod and the horizontal, which starts at \(60^\circ\).

The rod is released from rest, and we need to find the angular speed when the emf is maximum. When the rod is released, gravitational potential energy is converted into rotational kinetic energy.

Initially, the potential energy is:

\(PE = mg\left(\frac{\ell}{2}\right)\sin{60^\circ} = mg\left(\frac{\ell \sqrt{3}}{4}\right)\)

At the lowest point, potential energy is converted to kinetic energy:

\(\frac{1}{2} I \omega^2\)

The moment of inertia \((I)\) about the hinge point is:

\(I = \frac{1}{3} m \ell^2\)

Using conservation of energy:

\(mg\left(\frac{\ell \sqrt{3}}{4}\right) = \frac{1}{2} \left(\frac{1}{3} m \ell^2\right) \omega^2\)

Solving for \(\omega\):

\(\omega^2 = \frac{3g\ell\sqrt{3}}{2\ell^2}\)

\(\omega = \sqrt{\frac{3g}{2\ell}}\)

Substitute \(\omega\) into the emf formula:

\(\text{emf}_{max} = \frac{1}{2} B \ell^2 \sqrt{\frac{3g}{2\ell}} \sin{90^\circ}\)

\(\text{emf}_{max} = B \ell \sqrt{\frac{3g\ell}{8}}\)

Thus, the maximum induced emf in the rod is \(B \ell \sqrt{\frac{3g\ell}{8}}\), which corresponds to the given correct option.

Therefore, the correct answer is:

$B\ell\sqrt{\dfrac{3g\ell}{8}}$