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Chapter 14: Problem 2039

A rectangular loop with a sliding rod of length \(2 \mathrm{~m} \&\) resistance \(2 \Omega\). It moves in a uniform magnetic field of \(3 \mathrm{~T}\) perpendicular to plane of loop. The external force required to keep the rod moving with constant velocity of \(2 \mathrm{~ms}^{-1}\) is

Short Answer

Expert verified
The external force required to keep the sliding rod moving at a constant velocity of \(2 \mathrm{ms}^{-1}\) in the given magnetic field is \(36 \mathrm{N}\).

Step by step solution

01

Determine magnetic flux

First, we need to determine the magnetic flux through the rectangular loop. The magnetic flux, denoted by \(\Phi_B\), is given by: \[\Phi_B = B \cdot A\] where \(B\) is the uniform magnetic field and \(A\) is the area of the loop. Since the magnetic field is perpendicular to the plane of the loop, the magnetic flux becomes: \[\Phi_B = B \cdot A \cdot \cos{0} = B \cdot A\]
02

Calculate the induced EMF

As the rod moves with a constant velocity, the area of the loop changes over time. This change in the magnetic flux induces an electromotive force (EMF). Let's denote the length of the sliding rod as \(l\) and its velocity as \(v\). According to Faraday's law of electromagnetic induction, the induced EMF (\(E_{ind}\)) is given by: \[E_{ind} = \frac{d\Phi_B}{dt} = B \frac{dA}{dt}\] We know that the width of the loop is changing at a constant velocity, so: \[\frac{dA}{dt} = l \cdot \frac{dl}{dt} = l \cdot v\] Thus, the induced EMF can be expressed as: \[E_{ind} = B \cdot l \cdot v\] Using given values, \(B = 3 T\), \(l = 2 m\), and \(v = 2 m/s\), we get: \[E_{ind} = 3 \cdot 2 \cdot 2 = 12 \mathrm{V}\]
03

Determine the induced current

Now, we will find the induced current using Ohm's law. The resistance of the sliding rod is given as \(R = 2 \Omega\). According to Ohm's law, we have: \(I = \frac{E_{ind}}{R}\) Substituting the values, we get \[I = \frac{12}{2} = 6 \mathrm{A}\]
04

Calculate the magnetic force on the sliding rod

Due to the current, the magnetic field exerts a force on the sliding rod. This magnetic force, denoted by \(F_B\), is given by the expression: \[F_B = I \cdot l \cdot B\] Using the calculated current and given length and magnetic field values, we obtain: \[F_B = 6 \cdot 2 \cdot 3 = 36 \mathrm{N}\]
05

Determine the external force

Finally, to keep the rod moving at a constant velocity in the presence of the magnetic force, an external force must counterbalance that magnetic force. According to Newton's second law: \[\sum F = F_{external} - F_B = 0\] Thus, the required external force is: \[F_{external} = F_B = 36 \mathrm{N}\] The external force required to keep the sliding rod moving at a constant velocity of 2 m/s in the given magnetic field is 36 N.

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Most popular questions from this chapter

The instantaneous voltage through a device of impedance \(20 \Omega\) is \(\varepsilon=80 \sin 100 \pi t\). The effective value of the current is, (a) \(3 \mathrm{~A}\) (b) \(2.828 \mathrm{~A}\) (c) \(1.732 \mathrm{~A}\) (d) \(4 \mathrm{~A}\)

An LCR series circuit with \(\mathrm{R}=100 \Omega\) is connected to a $200 \mathrm{~V}\(, \)50 \mathrm{~Hz}$ a.c source when only the capacitance is removed the current lies the voltage by \(60^{\circ}\) when only the inductance is removed, the current leads the voltage by \(60^{\circ}\). The current in the circuit is, (a) \(2 \mathrm{~A}\) (b) \(1 \mathrm{~A}\) (c) \((\sqrt{3} / 2) \mathrm{A}\) (d) \((2 / \sqrt{3}) \mathrm{A}\)

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