Question

Consider an annular region in free space containing a uniform magnetic field in the \(z\)-direction, schematically represented by the shaded region in the figure. A particle having charge \(Q\) and mass \(M\) starts from point \(P(a, 0, 0)\) in the \(+x\)-direction with constant speed \(v\). If the radii of inner and outer circles are \(a\) and \(b\), respectively, the minimum magnetic field required so that the particle returns to the inner circle is: 

• A. \(\dfrac{Mv}{Q} \left( \dfrac{b^2 - a^2}{b} \right)^{-1}\)
• B. \(\dfrac{Mv}{Q} \left( \dfrac{b^2 - a^2}{2b} \right)^{-1}\)
• C. \(\dfrac{Mv}{Q} \left( \dfrac{b^2 - a^2}{3b} \right)^{-1}\)
• D. \(\dfrac{Mv}{Q} \left( \dfrac{b^2 - a^2}{4b} \right)^{-1}\)

Hint:

For circular motion of a charged particle in a magnetic field, always equate the magnetic force \(qvB\) to the centripetal force \(mv^2/r\).

Answer 1

The Correct Option is B

Step 1: Concept of circular motion in a magnetic field. 
The radius of circular motion of a charged particle moving in a magnetic field \(B\) is \[ r = \frac{Mv}{QB} \]

Step 2: Condition for returning to the inner circle. 
The particle must just touch the outer boundary before returning to the starting point, forming a semicircular trajectory between \(a\) and \(b\). At the maximum deflection, the center of circular motion lies along the \(x\)-axis. The geometry gives: \[ (b - a) = 2r \]

Step 3: Substitute radius. 
\[ r = \frac{b^2 - a^2}{2b} = \frac{Mv}{QB} \Rightarrow B = \frac{Mv}{Q} \left( \frac{b^2 - a^2}{2b} \right)^{-1} \]

Step 4: Conclusion. 
The minimum magnetic field required is \(\dfrac{Mv}{Q} \left( \dfrac{b^2 - a^2}{2b} \right)^{-1}\).