Q4. Assertion: a metal rod moving through a uniform magnetic field can have an emf across it even though the field does not change. Reason: the Lorentz force on the free charges in a moving rod pushes them along it. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: for mutually perpendicular rod, velocity and field, integrating the motional term v × B along the rod gives Blv. The energy comes from mechanical work maintaining the motion, not net work by magnetic force (NCERT page 163).
Q5. A metal rod 1.0 m long turns at 400 rad/s about an axis through one end, normal to the rod, with its other end touching a metal ring. A uniform field of 0.5 T is parallel to the axis. Find the emf between the centre and the ring. (NCERT Exercise 6.5)
ε = ½Bωl2 = ½ × 0.5 × 400 × 1.02 = 100 V (NCERT Example 6.6, page 164).
Q6. A rod 0.50 m long slides at a steady 2.0 m/s on rails in a field of 0.40 T normal to the plane of the rails, and the circuit has a resistance of 4.0 Ω. Find the emf, the current, the force needed to keep the rod moving and the power supplied.
ε = Blv = 0.40 × 0.50 × 2.0 = 0.40 V, so I = 0.40/4.0 = 0.10 A. The opposing force is IlB = 0.10 × 0.50 × 0.40 = 0.020 N, so the outside force must be 0.020 N, and its power is Fv = 0.020 × 2.0 = 0.040 W, equal to I2R = 0.102 × 4.0 = 0.040 W (beyond NCERT, reasoned from NCERT Eqs. (4.4) and (6.5)).
Q7. An aeroplane with a wingspan of 30 m flies horizontally at 200 m/s where the vertical component of the earth’s field is 4.0 × 10−5 T. Find the emf between its wing tips.
The wings, the velocity and the vertical field are mutually perpendicular, so ε = Blv = 4.0 × 10−5 × 30 × 200 = 0.24 V (NCERT Eq. (6.5), page 162).