Question

A metallic rod of 10 cm is rotated with a frequency 100 revolution per second about an axis perpendicular to its length and passing through its one end in a uniform transverse magnetic field of strength 1 T. The emf developed across its ends is:

• A. 628 V
• B. 3.14 V
• C. 31.4 V
• D. 6.28 V

Hint:

The emf induced in a rotating conductor in a magnetic field can be found using the formula \( \text{emf} = \frac{1}{2} B \omega L^2 \), where \( \omega \) is the angular velocity and \( L \) is the length of the ro(D)

Answer 1

The Correct Option is B

The emf induced in a rotating rod in a magnetic field is given by the formula: \[ \text{emf} = \frac{1}{2} B \omega L^2 \] where: - \( B \) is the magnetic field strength, - \( \omega \) is the angular velocity of the rod, - \( L \) is the length of the ro(D) First, let's calculate the angular velocity \( \omega \): - The frequency of rotation is \( f = 100 \, \text{rev/s} \), - The angular velocity is \( \omega = 2 \pi f = 2 \pi \times 100 = 200 \pi \, \text{rad/s} \). The length of the rod is \( L = 10 \, \text{cm} = 0.1 \, \text{m} \), and the magnetic field strength is \( B = 1 \, \text{T} \). Now, substitute these values into the formula for emf: \[ \text{emf} = \frac{1}{2} \times 1 \times 200 \pi \times (0.1)^2 \] \[ \text{emf} = \frac{1}{2} \times 200 \pi \times 0.01 \] \[ \text{emf} = 3.14 \, \text{V} \] Thus, the induced emf across the ends of the rod is \( 3.14 \, \text{V} \).