Q4. Assertion: a magnetic field cannot speed up or slow down a charged particle. Reason: the magnetic force on a moving charge is always at right angles to its velocity. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: a force at right angles to the velocity does no work, so the kinetic energy and the speed stay the same (NCERT Summary point 1, page 132).
Q5. What is the magnetic force per unit length on a wire carrying 8 A at 30° to a uniform field of 0.15 T? (NCERT Exercise 4.5)
F/l = IB sin θ = 8 × 0.15 × 0.5 = 0.6 N m−1 (NCERT Eq. (4.4), page 110).
Q6. A 3.0 cm wire carrying 10 A is placed inside a solenoid, at right angles to its axis. The field inside is 0.27 T. Find the force on the wire. (NCERT Exercise 4.6)
The field inside a solenoid is along its axis, so the wire is at right angles to it: F = IlB = 10 × 0.030 × 0.27 ≈ 8.1 × 10−2 N, at right angles to both the wire and the axis (NCERT Eq. (4.4), page 110).
Q7. A proton passes straight through a region where an electric field of 2.0 × 104 V/m and a magnetic field of 0.10 T are at right angles to each other and to its velocity. Find its speed.
It is undeflected when the electric and magnetic forces cancel: qE = qvB, so v = E/B = 2.0 × 104/0.10 = 2.0 × 105 m/s (reasoned from NCERT Eq. (4.3); this arrangement, often called a velocity selector, is not in the current NCERT book). The speed picked out, E/B, is the same for any charge and mass, so an electron or an alpha particle at this speed also goes straight through.
Q8. A charge of 2.0 μC moves with velocity v = (3î + 4ĵ) × 105 m/s in a field B = 0.50k̂ T. Find the magnetic force.
î × k̂ = −ĵ and ĵ × k̂ = î. So v × B = 0.50 × 105 × (4î − 3ĵ) = (2.0î − 1.5ĵ) × 105. Then F = 2.0 × 10−6 × (2.0î − 1.5ĵ) × 105 = (0.40î − 0.30ĵ) N, of size 0.50 N, at right angles to both v and B (reasoned from NCERT Eq. (4.3)).