Question

An ideal solenoid of cross-sectional area ${10}^{-4}{m}^2$ has 500 turns per metre. At the centre of this solenoid, another coil of 100 turns is wrapped closely around it. If the current in the coil changes from 0 to 2 A in 3.14 millisecond, the emf developed in the second coil is

• A. 1 mV
• B. 2 mV
• C. 3 mV
• D. 4 mV

Answer 1

The correct option is D

4 mV

The magnetic field due to the solenoid is
$B={\mu }_{0}nI$
where n is the number of turns per unit length. Since the second coil is wrapped closely around the solenoid, the cross-sectional area of the coil can be taken to be equal to that of the solenoid. Since the initial current and hence the initial magnetic field is zero, the change of flux for a single turn is ${\mu }_{0}nIA$, where A is the cross-sectional area. Therefore, induced emf for a single turn is
$|e|=\frac{{\mu }_{0}nIA}{t}=\frac{4\times 3.14\times {10}^{-7}\times 500\times 2\times {10}^{-4}}{3.14\times {10}^{-3}}=4\times {10}^{-5}V\therefore Inducedemffor100turns=4\times {10}^{-5}\times 100=4\times {10}^{-3}=4mV.$