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Chapter 14: Problem 2027

Two co-axial solenoids are made by a pipe of cross sectional area $10 \mathrm{~cm}^{2}\( and length \)20 \mathrm{~cm}$ If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is...... (a) \(4.8 \pi \times 10^{-4} \mathrm{H}\) (b) \(4.8 \pi \times 10^{-5} \mathrm{H}\) (c) \(2.4 \pi \times 10^{-4} \mathrm{H}\) (d) \(4.8 \pi \times 10^{4} \mathrm{H}\)

Short Answer

Expert verified
The mutual inductance between the two co-axial solenoids is (b) \(4.8π × 10^{-5} \, \mathrm{H}\).

Step by step solution

01

Calculate the self-inductance of each solenoid

First, we need to find the self-inductance L1 and L2 of the solenoids using the formula: \[L = \frac{μ₀ * A * n^2}{l}\] where μ₀ = permeability of free space = \(4π × 10^{-7} \, \mathrm{T \, m/A}\), A is the cross-sectional area of the pipe, n is the number of turns per unit length and l is the length of the solenoid. For solenoid 1 \(n_1 = \frac{300}{0.2} \, \mathrm{turns/m}\) For solenoid 2 \(n_2 = \frac{400}{0.2} \, \mathrm{turns/m}\)
02

Calculate the mutual inductance between the solenoids

To find the mutual inductance M, we use the following formula: \[M = \frac{μ₀ * A * n_1 * n_2}{l}\] Insert the values and perform the calculation: \(M = \frac{4π × 10^{-7} * 10^{-2} * (\frac{300}{0.2}) * (\frac{400}{0.2})}{0.2}\)
03

Calculate the answer

Now, simplify and solve the expression for M: \(M = 4.8π × 10^{-5} \, \mathrm{H}\) So the answer is (b) \(4.8π × 10^{-5} \, \mathrm{H}\).

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Most popular questions from this chapter

A magnetic field \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(200 \mathrm{~cm}^{2}\) with 25 turns. The average emf induced in the coil is \(0.1 \mathrm{v}\) when it removes from the field in time t. The value of \(\mathrm{t}\) is (a) \(0.1 \mathrm{sec}\) (b) \(1 \mathrm{sec}\) (c) \(0.01 \mathrm{sec}\) (d) \(20 \mathrm{sec}\)

\(\mathrm{A}\) coil of inductance \(\mathrm{L}\) has an inductive reactance of \(\mathrm{X}_{\mathrm{L}}\) in an AC circuit in which the effective current is \(\mathrm{I}\). The coil is made from a super-conducting material and has no resistance. The rate at which power is dissipated in the coil is. (a) 0 (b) \(\mathrm{IX}_{\mathrm{L}}\) (c) \(I^{2} \mathrm{X}_{\mathrm{L}}\) (d) \(\mathrm{IX}^{2} \mathrm{~L}\)

A coil having n turns \(\&\) resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 \mathrm{R} \Omega\). This combination is moved from a magnetic field \(\mathrm{W}_{1} \mathrm{~Wb}\) to $\mathrm{W}_{2} \mathrm{~Wb}\( in \)\mathrm{t}$ second. The induced current in the circuit is.... (a) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rnt}\\}\right]$ (b) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rt}\\}\right]$ (c) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rnt}\\}\right]$ (d) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rt}\\}\right]$

A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V}\). The current reaches half of its steady state value in \(\ldots \ldots\) (a) \(0.15 \mathrm{sec}\) (b) \(0.3 \mathrm{sec}\) (c) \(0.05 \mathrm{sec}\) (d) \(0.1 \mathrm{sec}\)

The resistance of a coil for dc is in ohms. In ac, the resistance (a) will remain same (b) will increase (c) will decrease (d) will be zero
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