ElectroMagnetic InductionMedium
Question
A 30 cm long solenoid has 10 turns per cm and area of $5{\text{ }cm}^{2}$. The current through the solenoid coil varies from 2 A to 4 A is 3.14 s . The e.m.f. induced in the coil is $\alpha \times 10^{- 5}\text{ }V$. The value $\alpha$ is $\_\_\_\_$ .
Options
A.60
B.12
C.120
D.34
Solution
Sol. [Key Concept: Solenoid self-inductance gives $|\varepsilon| = \mu_{0}n^{2}A\mathcal{l}\frac{di}{dt} = 12 \times 10^{- 5}$ V, so $\alpha = 12$.]
$${\varepsilon = - \frac{Ldi}{dt} = - \mu_{0}n^{2}\text{ }A\mathcal{l}\frac{di}{dt} }{= - 4\pi \times 10^{- 7} \times 10^{6} \times 5 \times 10^{- 4} \times 0.3 \times \frac{2}{\pi} }{\Rightarrow - 12 \times 10^{- 5} }$$
Create a free account to view solution
View Solution FreeMore ElectroMagnetic Induction Questions
A rectangular loop of sides a & b is placed in magnetic field B. The emf induced in coil when normal of coil makes angle...In the circuit shown in figure what is the value of I1 just after pressing the key K ?...The induced currents always produce expanding magnetic fields round their conductors in a direction that opposes the ori...In the circuit shown, the cell is ideal. the coil has an inductance of 4H and zero resistance. F is a fuse of zero resis...A circular current loop of radius R is placed inside square loop of side length $L(L \gg R)$ such that they are co-plana...