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Chapter 16: Problem 22

A capacitance \(C\) is connected to a conducting rod of length \(\ell\) moving with a velocity \(v\) in a transverse magnetic field \(B\) then the charge developed in the capacitor is (A) Zero (B) \(B \ell v C\) (C) \(\frac{B l v C}{2}\) (D) \(\frac{B l v C}{3}\)

Short Answer

Expert verified
The short answer is: The charge developed in the capacitor is (B) \(B \ell v C\).

Step by step solution

01

Faraday's law of electromagnetic induction states that when a conducting loop moves through a magnetic field, an electromotive force (EMF) is induced in the loop. The magnitude of the EMF is proportional to the rate of change of magnetic flux through the loop. The formula for the EMF induced in a straight rod of length l, moving with velocity v in a magnetic field B perpendicular to the rod, is given by: EMF = BLv where B is the magnetic field, L is the length of the rod, and v is the velocity of the rod. #Step 2: Relate the EMF of the rod to the voltage across the capacitor#

Since the rod is connected to the capacitor, the EMF induced across the rod due to the magnetic field is equal to the voltage across the capacitor (V). Therefore: V = BLv #Step 3: Determine the charge in the capacitor using capacitance formula#
02

The charge (Q) developed in the capacitor is related to the voltage across the capacitor and the capacitance according to the formula: Q = CV So, by substituting the expression for the voltage from step 2, we get: Q = C(BLv) #Step 4: Finding the correct answer choice#

Now, we can compare the expression for the charge developed in the capacitor (Q = C(BLv)) with the given answer choices: (A) Zero (B) \(B \ell v C\) (C) \(\frac{B \ell v C}{2}\) (D) \(\frac{B \ell v C}{3}\) The correct answer matches our expression for the charge developed in the capacitor, and it is: (B) \(B \ell v C\)

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Most popular questions from this chapter

A coil of \(20 \times 20 \mathrm{~cm}\) having 30 turns is making \(30 \mathrm{rps}\) in a magnetic field of 1 tesla. The peak value of the induced EMF is approximately (A) \(452 \mathrm{~V}\) (B) \(226 \mathrm{~V}\) (C) \(113 \mathrm{~V}\) (D) \(339 \mathrm{~V}\)

A conducting circular loop is placed in a uniform magnetic field of induction \(B\) tesla with its plane normal to the field. Now the radius of the loop starts shrinking at the rate \((d r / d t)\). Then the induced EMF at the instant when the radius is \(r\) will be (A) \(\pi r B\left(\frac{d r}{d t}\right)\) (B) \(2 \pi r B\left(\frac{d r}{d t}\right)\) (C) \(\pi r^{2}\left(\frac{d B}{d t}\right)\) (D) \(B \frac{\pi r^{2}}{2} \frac{d r}{d t}\)

A conducting rod of length \(2 \ell\) is rotating with constant angular speed \(\omega\) about its perpendicular bisector. A uniform magnetic field \(\vec{B}\) exists parallel to the axis of rotation. The EMF induced between two ends of the rod is (A) \(B \omega \ell^{2}\) (B) \(\frac{1}{2} B \omega \ell^{2}\) (C) \(\frac{1}{8} B \omega \ell^{2}\) (D) Zero

A positive charge is passing through an electromagnetic field in which \(\vec{E}\) and \(\vec{B}\) are directed towards \(y\)-axis and \(z\)-axis, respectively. If a charged particle passes through the region undeviated, then its velocity is/are represented by (here \(a, b\), and \(c\) are constant) (A) \(\vec{v}=\frac{E}{B} \hat{i}+a \hat{j}\) (B) \(\vec{v}=\frac{E}{B} \hat{i}+b \hat{k}\) (C) \(\vec{v}=\frac{E}{B} \hat{i}+c \hat{i}\) (D) \(\vec{v}=\frac{E}{B} \hat{i}\)

A magnetic field \(\vec{B}=\left(\frac{B_{0} y}{a}\right) \hat{k}\) is into the paper in the \(+z\) direction. \(B_{0}\) and \(a\) are positive constants. A square loop \(E F G H\) of side \(a\), mass \(m\), and resistance \(R\), in \(x-y\) plane, starts falling under the influence of gravity. Assume \(x\)-axis is horizontal and \(y\) is vertically downward. The magnitude and direction of the induced current in the loop when its speed is \(v\) is (A) \(\frac{B_{0} a v}{R}\), anti-clockwise (B) \(\frac{B_{0} a v}{R}\), clockwise (C) \(\frac{B_{0} v}{a R}\), anti-clockwise (D) None of these
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