Q4. Assertion: the field inside a long solenoid is the same near its wall as on its axis. Reason: B = μnI does not involve the distance from the axis. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: Ampere’s law gives B = μnI for any position of the side ab inside the solenoid, so the field is uniform across it (NCERT Eq. (4.16), page 122).
Q5. A long solenoid of 1000 turns per metre carries 3.0 A. A second, coaxial solenoid inside it has 500 turns per metre and carries 2.0 A the opposite way. Find the field inside the inner solenoid.
The fields add as vectors: 4π × 10−7 × (1000 × 3.0 − 500 × 2.0) = 4π × 10−7 × 2000 ≈ 2.5 × 10−3 T, in the direction set by the outer solenoid’s current (NCERT Eq. (4.16) and superposition, page 109).
Q6. A solenoid is wound with a single layer of insulated wire of outside diameter 0.50 mm, adjacent insulated turns touching, and carries 1.5 A. Find the field inside.
Touching turns of 0.50 mm wire give n = 1/(0.50 × 10−3) = 2000 per metre, so B = 4π × 10−7 × 2000 × 1.5 ≈ 3.8 × 10−3 T (NCERT Eq. (4.16), page 122).
Q7. A wire of fixed length is wound into a long solenoid of fixed length, first on a tube of radius r and then on a tube of radius 2r, and carries the same current both times. How does the field inside change?
Neglecting helical pitch and lead length, and assuming both windings remain long, on the wider tube each turn uses twice as much wire, so there are half as many turns: n halves and B = μnI halves (reasoned from NCERT Eq. (4.16)).