Question

A long solenoid having 200 turns per cm carries a current of 1.5 amp. At the centre of it is placed a coil of 100 turns of cross-sectional area 3.14 × 10⁻⁴ m² having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 sec, the induced e.m.f. in the coil is:

पर्याय

• 0.48 V

• 0.048 V

• 0.0048 V

• 48 V

Answer 1

0.048 V

Explanation:

B = μ₀ ni = (4π × 10⁻⁷) (200 × 10⁻²) × 1.5

= 3.8 × 10⁻² Wb/m²

Magnetic flux through each turn of the coil

Φ = BA = (3.8 × 10⁻²) (3.14 × 10⁻⁴) = 1.2 × 10⁻⁵ weber

When the current in the solenoid is reversed, the change in magnetic flux

= 2 × (1.2 × 10⁻⁵) = 2.4 × 10⁻⁵ weber

Induced e.m.f. = `"N"("d"Φ)/"dt"`

= `100 xx (2.4 xx 10^-5)/0.05`

= 0.048 V