Question

A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ the speed of ring is V and emf developed across the ring is

• A. zero
• B. \(2RBV\) and Q is at higher potential
• C. \(RBV\) and Q is at higher potential
• D. \(\frac{BV\pi R^{2}}{2}\) and M is at higher potential

Hint:

For a curved conductor in a uniform field, the emf equals that of the straight line joining its ends.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a conductor of any shape moving in a uniform field, the motional emf equals that of the straight line joining its ends, \(\varepsilon=B\,l_{eff}\,v\).

Step 2: Effective length
The ends M and Q of the semicircle are joined by a diameter of length \(2R\), perpendicular to both \(\vec{v}\) and \(\vec{B}\).

Step 3: Magnitude
\(\varepsilon=B(2R)V=2RBV\)

Step 4: Polarity
The magnetic force on positive charges is along \(\vec{v}\times\vec{B}\). With \(\vec{v}\) downward and \(\vec{B}\) into the page this points towards Q, so Q is at higher potential. This is option (B).

Final Answer:
The emf is \(2RBV\), with Q at the higher potential. \[ \boxed{\text{(B) }2RBV\text{ and Q is at higher potential}} \]