Q4. Assertion: the magnetic flux through a loop can be negative. Reason: ΦB = BA cos θ, and cos θ is negative when the angle between B and the normal is more than 90°. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: flux is a scalar whose sign depends on the angle between B and the chosen normal (NCERT Eq. (6.1), page 157).
Q5. Name three ways of changing the magnetic flux through a coil placed in a magnetic field.
Change the field B (for example by moving a magnet nearer), change the area A (by shrinking or stretching the coil), or change the angle θ (by rotating the coil) (NCERT page 158).
Q6. A horizontal loop of area 0.50 m2 lies where the earth’s field is 5.0 × 10−5 T, directed at 30° below the horizontal. Find the magnitude of the flux through the loop.
Taking the normal downward, it makes 90° − 30° = 60° with the field. ΦB = 5.0 × 10−5 × 0.50 × cos 60° = 1.25 × 10−5 ≈ 1.3 × 10−5 Wb (an upward normal gives −1.3 × 10−5 Wb; NCERT Eq. (6.1), page 157).
Q7. Show that 1 Wb = 1 V s.
Faraday’s law gives an emf as a rate of change of flux, ε = −dΦB/dt, so a volt is a weber per second and 1 Wb = 1 V s (reasoned from NCERT Eq. (6.3), page 158). In base units both are kg m2 s−2 A−1.
Q8. A long straight wire carries 10 A. A rectangular loop lies in the same plane, with its 0.20 m sides parallel to the wire at distances 0.10 m and 0.30 m from it. Find the magnitude of the flux through the loop.
The field of a long wire, B = μ0I/2πr (NCERT Eq. (4.14), page 118), is normal to the loop but falls with r. So add the flux through thin strips of width dr and area a dr, with a = 0.20 m: ΦB = ∫ (μ0Ia/2πr) dr from 0.10 m to 0.30 m = (μ0Ia/2π) ln 3 = 2 × 10−7 × 10 × 0.20 × 1.10 ≈ 4.4 × 10−7 Wb (the integral is beyond NCERT).