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Question: A solenoid has 2000 turns wound over a length of 0.314 m. Around its central section a coil of 100 t...

A solenoid has 2000 turns wound over a length of 0.314 m. Around its central section a coil of 100 turns and area of cross-section 1 × 10–3m2 is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 sec, the emf induced in the coil is equal to

A

6 × 10–4V

B

12.8 Mv

C

6 × 10–2V

D

12.8 V

Answer

12.8 Mv

Explanation

Solution

Magnetic field at the centre of the solenoid is given by B=μ0ni=μ0Nil=4×3.14×10−7×20000.314×2B = \mu_{0}ni = \frac{\mu_{0}Ni}{l} = 4 \times 3.14 \times 10^{- 7} \times \frac{2000}{0.314} \times 2 =16×10−3T= 16 \times 10^{- 3}T. This magnetic field is perpendicular to the plane of the coil

∴ Magnetic flux linked with coil φ=N′BA\varphi = N^{'}BA

∴ Induced emf e=−dφdt=−ddt(N′BA)=−N′AdBdt=−N′A(−B−B)dt=2N′BAdte = \frac{- d\varphi}{dt} = - \frac{d}{dt}(N^{'}BA) = - N^{'}A\frac{dB}{dt} = - N^{'}A\frac{( - B - B)}{dt} = \frac{2N^{'}BA}{dt}

∴ e=2×100×16×10−3×1×10−30.25=12.8mV.e = \frac{2 \times 100 \times 16 \times 10^{- 3} \times 1 \times 10^{- 3}}{0.25} = 12.8mV.