Question

A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

Hint:

For a rod rotating about one end in a magnetic field perpendicular to the plane of rotation, \[ e=\frac{1}{2}B\omega l^2 \] Always convert rpm into rad/s using \[ \omega=\frac{2\pi N}{60}. \]

Answer 1

Concept: When a conducting rod rotates in a magnetic field, the free charges present inside the rod experience a magnetic Lorentz force. This force causes charge separation along the length of the rod and an emf is induced between its ends. For a rod of length \(l\) rotating with angular velocity \(\omega\) in a uniform magnetic field \(B\) perpendicular to the plane of rotation, the induced emf is \[ \boxed{ e=\frac{1}{2}B\omega l^2 } \]

Step 1: Write the given quantities. Length of rod \[ l=50\ \text{cm} =0.50\ \text{m} \] Magnetic field \[ B=4.0\ \text{mT} =4.0\times10^{-3}\ \text{T} \] Angular speed \[ 60\ \text{rpm} \] \[ = 60\ \text{revolutions per minute} \] \[ = 1\ \text{revolution per second} \] Therefore, \[ f=1\ \text{Hz} \] Angular velocity \[ \omega=2\pi f \] \[ \omega=2\pi(1) \] \[ \boxed{ \omega=2\pi\ \text{rad s}^{-1} } \]

Step 2: Apply the expression for motional emf. \[ e = \frac{1}{2} B\omega l^2 \] Substituting the values, \[ e = \frac{1}{2} (4\times10^{-3}) (2\pi) (0.50)^2. \]

Step 3: Simplify the numerical calculation. Since \[ (0.50)^2=0.25, \] \[ e = \frac{1}{2} \times 4\times10^{-3} \times 2\pi \times 0.25. \] \[ e = \pi\times10^{-3}. \] \[ e = 3.14\times10^{-3}\ \text{V}. \] Therefore, \[ \boxed{ e=3.14\times10^{-3}\ \text{V} } \]

Step 4: Express the answer in millivolts. \[ e = 3.14\ \text{mV}. \] \[ \boxed{ e=3.14\ \text{mV} } \] Final Answer: \[ \boxed{ e=3.14\times10^{-3}\ \text{V} } \] or \[ \boxed{ e=3.14\ \text{mV} } \]