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Chapter 13: Problem 1907

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge $1.6 \times 10^{-19} \mathrm{C}\( and moving with a velocity of \)10^{6} \mathrm{~m} / \mathrm{s}$ enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

Short Answer

Expert verified
The magnetic field intensity must be approximately \(5.6 \times 10^{-4} T\), which is closest to option (b) \(5.6 \times \overline{10^{-5}} T\) or \(5.6 \times 10^{-4}T\).

Step by step solution

01

Note down the given variables

The problem provides the following information: - Mass of the electron, m = \(9 \times 10^{-31} kg\) - Charge of the electron, q = \(1.6 \times 10^{-19} C\) - Velocity of the electron, v = \(10^{6} m/s\) - Radius of the circular path, r = \(0.10 m\)
02

Equate the magnetic force to the centripetal force

Since the magnetic force is acting as a centripetal force in this case, we can equate the two forces: \(q * v * B * sin \theta = m * v^2 / r\) Since the electron describes a circle, it indicates that the angle, \(\theta\) between the velocity and the magnetic field is \(90^{\circ}\). Therefore, \(sin \theta = 1\).
03

Solve for the magnetic field intensity, B

Now, we need to solve the equation for the magnetic field intensity, B. First, we will cancel out the v terms: \(q * B = m * v / r\) Now, we can solve for B: \(B = (m * v) / (q * r)\)
04

Substitute the given values and calculate the magnetic field intensity

Plug in the given values and the magnetic field intensity becomes: \(B = \frac{(9 \times 10^{-31} kg) * (10^{6} m/s)}{(1.6 \times 10^{-19} C) * (0.10 m)}\) By simplifying the expression: \(B = \frac{9 \times 10^5}{1.6 \times 10^{-1}} T\) \(B = \frac{9}{1.6} \times 10^{6-1} T\) \( B = 5.625 \times 10^{-4} T\)
05

Compare the calculated value with the given options

Our calculated value for the magnetic field intensity is \(5.625 \times 10^{-4} T\). Comparing this with the choices provided, we notice that none of the options exactly match this value. The closest option is (b) \(5.6 \times 10^{-5}T\), however, it carries an overlined digit (which means that the digit is repetetive). If we factor in that factor of 10 difference by considering this number as \(5.6 \times 10 \times 10^{-5}T\), the closest option would be: Option (b) \(5.6 \times \overline{10^{-5}} T\) or \(5.6 \times 10^{-4}T\) Thus, the correct (closest) answer is option (b).

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Most popular questions from this chapter

The magnetic induction at a point \(P\) which is at a distance \(4 \mathrm{~cm}\) from a long current carrying wire is \(10^{-8}\) tesla. The field of induction at a distance \(12 \mathrm{~cm}\) from the same current would be tesla. (a) \(3.33 \times 10^{-9}\) (b) \(1.11 \times 10^{-4}\) (c) \(3 \times 10^{-3}\) (d) \(9 \times 10^{-2}\)

A magnetic field existing in a region is given by $\mathrm{B}^{-}=\mathrm{B}_{0}[1+(\mathrm{x} / \ell)] \mathrm{k} \wedge . \mathrm{A}\( square loop of side \)\ell$ and carrying current I is placed with edges (sides) parallel to \(\mathrm{X}-\mathrm{Y}\) axis. The magnitude of the net magnetic force experienced by the Loop is (a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) (b) \((1 / 2) B_{0} I \ell\) (c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) (d) BI\ell

The magnetic susceptibility is negative for (a) Paramagnetic materials (b) Diamagnetic materials (c) Ferromagnetic materials (d) Paramagnetic and ferromagnetic materials

If a magnet of pole strength \(\mathrm{m}\) is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be (a) \((\mathrm{m} / 4)\) (b) \((\mathrm{m} / 2)\) (c) \((\mathrm{m} / 8)\) (d) \(4 \mathrm{~m}\)

A proton and an particle are projected with the same kinetic energy at right angles to the uniform mag. field. Which one of the following statements will be true. (a) The \(\alpha\) - particle will be bent in a circular path with a small radius that for the proton. (b) The radius of the path of the \(\alpha\) - particle will be greater than that of the proton. (c) The \(\alpha\) - particle and the proton will be bent in a circular path with the same radius. (d) The \(\alpha\) - particle and the proton will go through the field in a straight line.
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