Question

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B as shown in the figure. The potential difference developed across the ring when its speed is v, is
\(\{\sin (90)^{\circ }=1\}\)

• A. \(\pi rBv\) and terminal R is at higher potential
• B. \(\frac{Bv\pi r^{2}}{2}\) and terminal P is at higher potential
• C. \(2rBv\) and terminal R is at higher potential
• D. zero

Hint:

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

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The emf of a curved conductor moving in a uniform field equals that of the straight line joining its ends.

Step 2: Calculation
PR is the horizontal diameter, length \(2r\). The velocity is downward, the field is into the page and the diameter is horizontal, all mutually perpendicular.
\(e=B\,v\,(2r)=2rBv\)

Step 3: Polarity
The force on positive charges, \(\vec{v}\times\vec{B}\) (down \(\times\) into the page), points towards R. So R is at the higher potential.

Final Answer:
The potential difference is \(2rBv\) with R at the higher potential. \[ \boxed{\text{(C) }2rBv\text{ and terminal R is at higher potential}} \]