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Chapter 16: Problem 6

There is a current of \(1.344 \mathrm{~A}\) in a copper wire whose area of cross-section normal to the length of the wire is \(1 \mathrm{~mm}^{2}\). If the number of free electrons per \(\mathrm{cm}^{3}\) is \(8.4 \times 10^{22}\), then the drift velocity of electrons will be (A) \(1.0 \mathrm{~mm} \mathrm{~s}\) (B) \(1.0\) meter s (C) \(0.1 \mathrm{~mm} \mathrm{~s}\) (D) \(0.01 \mathrm{~mm} \mathrm{~s}\)

Short Answer

Expert verified
The drift velocity of electrons is \(10^{-8} \mathrm{~mm} \mathrm{~s}\), which is not among the given options.

Step by step solution

01

Determine the Given Values

The given values in the exercise are: Current (I) = 1.344 A (Amperes) Area of cross-section (A) = 1 mm^2 = \(1 \times 10^{-6} \mathrm{m}^2\) (converted to meters^2) Number of free electrons per volume (n) = 8.4 x \(10^{22}\) cm^3 = \(8.4 \times 10^{28} \mathrm{m}^{-3}\) (converted to meters^-3) Elementary charge (e) = \(1.6 \times 10^{-19} \mathrm{C}\) (Coulombs)
02

Rearrange the Current Formula to Solve for Drift Velocity

The formula for current through a conducting wire is, I = n * A * e * v_d. We are given I, n, A and e and we need to find v_d (the drift velocity of electrons). So, we will rearrange the formula to solve for v_d: Drift velocity (v_d) = I/(n * A * e)
03

Plug in the Given Values and Calculate the Drift Velocity

Now, we will use the given values for I, n, A, and e, and plug them into the rearranged formula to find v_d: v_d = 1.344 A / \((8.4 \times 10^{28} \mathrm{m}^{-3}) \times (1 \times 10^{-6} \mathrm{m}^2) \times (1.6 \times 10^{-19} \mathrm{C}) \) v_d = \(\frac{1.344}{(8.4 \times 10^{28}) \times (1 \times 10^{-6}) \times (1.6 脳 10^{-19})}\) v_d = \(1.344 / (1.344 \times 10^{-11})\) v_d = \(10^{-11}\) ms
04

Convert the Drift Velocity

Now, we need to convert the drift velocity found in step 3 (ms) to millimeters per second (as given in the options): v_d = \(10^{-11}\) ms * \(10^{3}\) (conversion factor from meters to millimeters) v_d = \(10^{-8} \mathrm{~mm} \mathrm{~s}\)
05

Compare the Result with the Given Options and Find the Answer

The drift velocity calculated is \(10^{-8} \mathrm{~mm} \mathrm{~s}\), which is not in the given options. This means that none of the options provided are correct. The closest option is (D) \(0.01\mathrm{~mm} \mathrm{~s}\), but it is still not the right answer.

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Most popular questions from this chapter

A conducting square loop of side \(L\) and resistance \(R\) moves in its plane with a uniform velocity \(v\) perpendicular to one of its sides. A magnetic field \(B\), constant in space and time, pointing perpendicular and into the plane of the loop exists everywhere as shown in Fig. 16.38. The current induced in the loop is (A) \(B L v / R\) clockwise (B) \(B L v / R\) anti-clockwise (C) \(2 B L v / R\) anti-clockwise (D) Zero

If a coil of metal wire is kept stationary in a non-uniform magnetic field, (A) An EMF and current are both induced in the coil (B) A current but no EMF is induced in the coil (C) An EMF but no current is induced in the coil (D) Neither EMF nor current is induced in the coil

An equilateral triangular loop having a resistance \(R\) and length of each side \(\ell\) is placed in a magnetic field which is varying at \(\frac{d B}{d t}=1 \mathrm{~T} / \mathrm{S}\). The induced current in the loop will be (A) \(\frac{\sqrt{3}}{4} \frac{l^{2}}{R}\) (B) \(\frac{4}{\sqrt{3}} \frac{l^{2}}{R}\) (C) \(\frac{\sqrt{3}}{4} \frac{R}{l^{2}}\) (D) \(\frac{4}{\sqrt{3}} \frac{R}{l^{2}}\)

Flux \(\phi\) (in weber) in a closed circuit of resistance \(10 \Omega\) varies with time \(t\) (in seconds) according to the equation \(\phi=6 t^{2}-5 t+1\). The magnitude of the induced current in the circuit at \(t=0.25 \mathrm{~s}\) is (A) \(0.2 \mathrm{~A}\) (B) \(0.6 \mathrm{~A}\) (C) \(0.8 \mathrm{~A}\) (D) \(1.2 \mathrm{~A}\)

Net force on a current carrying loop kept in uniform magnetic field is zero and the torque on the loop \(\vec{\tau}=\vec{M} \times \vec{B}\), where \(M\) and \(B\) are magnetic dipole moment and magnetic field intensity, respectively. If it is free to rotate, then it will rotates about an axis passing through its centre of mass and parallel to \(\vec{\tau}\). Potential energy of the loop is given by \(U=-\vec{M} \cdot \vec{B}\). Assume a current carrying ring with its centre at the origin and having moment of inertia \(2 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}\) about an axis passing through one of its diameter and magnetic moment \(\vec{M}=(3 \hat{i}-4 \hat{j}) \mathrm{Am}^{2}\). At time \(t=0\), a magnetic field \(\vec{B}=(4 \hat{i}-3 \hat{j}) T\) is switched on. Then Torque acting on the loop is (A) Zero (B) \(25 \hat{k} \mathrm{Nm}\) (C) \(16 \hat{k} \mathrm{Nm}\) (D) \(10 \hat{k} \mathrm{Nm}\)
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