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Chapter 14: Problem 2020

When a train travels with a speed \(360 \mathrm{~km} \mathrm{~h}^{-1}\) along the track separated by 1 meter. The vertical component of earth's magnetic field is \(0.1 \times 10^{-4} \mathrm{~T}\). What is the value of induced emf between the rails? (a) \(10^{-2} \mathrm{~V}\) (b) \(10^{-4} \mathrm{~V}\) (c) \(10^{-3} \mathrm{~V}\) (d) \(1 \mathrm{~V}\)

Short Answer

Expert verified
The short answer is: (a) $10^{-2} \mathrm{~V}$

Step by step solution

01

To convert the velocity from km/h to m/s, multiply it by (1000 m/km) / (3600 s/h). \(v = 360 \frac{km}{h} \times \frac{1000m}{1km} \times \frac{1h}{3600s} \) #Step 2: Substitute values into the EMF formula#

Plug the values for B, v, and L into the EMF formula: EMF = B * v * L where B = 0.1 × 10^{-4} T, v = (calculated in step 1) and L = 1 m. #Step 3: Calculate induced EMF between the rails#
02

Now, calculate the induced EMF by multiplying the values as follows: EMF = (0.1 × 10^{-4} T) * (value obtained in step 1) * (1 m) #Step 4: Compare result with given options and select the correct one#

After calculating the EMF, compare the result with the given options (a) to (d) and select the one that matches the value you found.

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Most popular questions from this chapter

The variation of induced emf with time in a coil if a short bar magnet is moved along its axis with constant velocity is.

When 100 volt dc is applied across a coil, a current of \(1 \mathrm{~A}\) flows through it. When 100 volt ac at 50 cycle \(\mathrm{s}^{-1}\) is applied to the same coil, only \(0.5\) A current flows. The impedance of the coil is, (a) \(100 \Omega\) (b) \(200 \Omega\) (c) \(300 \Omega\) (d) \(400 \Omega\)

In an ac circuit the emf (e) and the current (i) at any instant core given respectively by \(\mathrm{e}=\mathrm{E}_{\mathrm{O}}\) sin $\operatorname{ct}, \mathrm{I}=\mathrm{I}_{\mathrm{O}} \sin (\cot -\Phi)$. The average power in the circuit over one cycle of ac is. (a) \(\left[\left\\{E_{O} I_{O}\right\\} / 2\right] \cos \Phi\) (b) \(\mathrm{E}_{\mathrm{O}} \mathrm{I}_{\mathrm{O}}\) (c) \(\left[\left\\{E_{Q} I_{Q}\right\\} / 2\right]\) (d) \(\left[\left\\{E_{O} I_{O}\right\\} / 2\right] \sin \Phi\)

An LC circuit contains a \(20 \mathrm{mH}\) inductor and a \(50 \mu \mathrm{F}\) capacitor with an initial charge of \(10 \mathrm{mc}\). The resistance of the circuit is negligible. At the instant the circuit is closed be \(t=0 .\) At what time is the energy stored completely magnetic. (a) \(\mathrm{t}=0 \mathrm{~ms}\) (b) \(\mathrm{t}=1.54 \mathrm{~ms}\) (c) \(\mathrm{t}=3.14 \mathrm{~ms}\) (d) \(\mathrm{t}=6.28 \mathrm{~ms}\)

An alternating voltage \(\mathrm{E}=200 \sqrt{2} \sin (100 \mathrm{t})\) is connected to 1 microfarad capacitor through an ac ammeter. The reading of the ammeter shall be. $\begin{array}{llll}\text { (a) } 10 \mathrm{~mA} & \text { (b) } 20 \mathrm{~mA} & \text { (c) } 40 \mathrm{~mA} & \text { (d) } 80 \mathrm{~mA}\end{array}$
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