Q4. Assertion: two identical charged particles moving at right angles to the same uniform magnetic field, at different speeds, take the same time to go round once. Reason: the radius of the path is proportional to the speed. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: the faster particle travels a longer path, 2πr, in proportion to its speed, so T = 2πr/v = 2πm/(qB) is the same (NCERT Eqs. (4.5) and [4.6(a)], page 112).
Q5. A uniform field of 6.5 G (1 G = 10−4 T) is kept in a chamber. An electron is shot into it at 4.8 × 106 m/s at right angles to the field. Explain why its path is a circle, and find the radius (me = 9.1 × 10−31 kg). (NCERT Exercise 4.11)
The magnetic force is at right angles to the velocity and has the constant size evB, so it acts as a centripetal force. r = mev/(eB) = 9.1 × 10−31 × 4.8 × 106/(1.6 × 10−19 × 6.5 × 10−4) ≈ 4.2 × 10−2 m = 4.2 cm (NCERT Eq. (4.5), page 112). NCERT’s exercise prints e = 1.5 × 10−19 C; the electron’s charge is 1.6 × 10−19 C, the value NCERT uses in Example 4.3 (page 113). With the printed value the answer would come out as about 4.5 cm.
Q6. In Q5, find the frequency of revolution of the electron. Does it depend on the speed? (NCERT Exercise 4.12)
ν = eB/(2πme) = 1.6 × 10−19 × 6.5 × 10−4/(2π × 9.1 × 10−31) ≈ 1.8 × 107 Hz = 18 MHz. It does not depend on the speed (NCERT Eq. [4.6(a)], page 112). With NCERT’s printed e = 1.5 × 10−19 C it would come out as about 17 MHz.
Q7. A proton and an alpha particle are accelerated from rest through the same potential difference and then enter the same magnetic field at right angles to it. Find the ratio of the radii of their paths.
Each gains kinetic energy K = qV, so r = √(2mK)/(qB) = √(2mV/q)/B ∝ √(m/q). For the alpha particle m/q is twice that of the proton, so rα/rp = √2 ≈ 1.4 (reasoned from NCERT Eq. (4.5)).
Q8. A proton enters a strip of uniform magnetic field of width d, moving at right angles both to the field and to the strip’s edge. In the field its path would be a circle of radius r = 2d. Through what angle is it turned, and for what fraction of a period is it in the field?
Inside the strip it moves on an arc of radius r whose centre lies on the entry edge. It leaves after crossing the width d, so sin θ = d/r = 1/2 and θ = 30°. The time is (30°/360°)T = T/12, and its speed is unchanged. If d were more than r, it would turn through 180° and leave through the edge it came in by (reasoned from NCERT Eq. (4.5)).