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 figure. The potential difference developed across the ring when its speed is v, is:

• A. \(\frac{Βνπr^2}{2}\) and P is at higher potential
• B. πrBv and R is at higher potential
• C. 2rBv and R is at higher potential
• D. Zero

Answer 1

The Correct Option is C

To solve this problem, we need to determine the potential difference developed across the semicircular conducting ring when it is moving in a magnetic field.

Given:

• Radius of the semicircular ring = \( r \)
• Magnetic field (\( B \)) is horizontal and perpendicular to the plane of the ring.
• Velocity of the ring (\( v \))

We will use the concept of motional EMF. When a conductor moves in a magnetic field, an electromotive force (EMF) is induced across its ends. The formula for EMF (\( \mathcal{E} \)) induced in a conductor of length \( L \) moving with velocity \( v \) perpendicular to a magnetic field \( B \) is given by:

\(\mathcal{E} = BvL\)

For a semicircular wire, the length of the wire from \( P \) to \( R \) through \( Q \) is \( \pi r \), but the effective length across which the EMF is induced is the straight line distance from \( P \) to \( R \), which is the diameter of the semicircle, i.e., \( 2r \).

Thus, the induced EMF (potential difference) is:

\(\mathcal{E} = Bv(2r) = 2rBv\)

Determining the polarity: As the ring falls, electrons in the segment PR will drift towards R due to the magnetic force, making R at a higher potential.

Therefore, the correct answer is: 2rBv and R is at higher potential.