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Chapter 27: Problem 25

A proton is accelerated from rest by a potential difference of \(400 .\) V. The proton enters a uniform magnetic field and follows a circular path of radius \(20.0 \mathrm{~cm} .\) Determine the magnitude of the magnetic field.

Short Answer

Expert verified
Answer: The magnitude of the magnetic field is approximately \(7.97 \times 10^{-4}\) T.

Step by step solution

01

Identify the known values

We are given: - The potential difference (V) = 400 V - The radius of the circular path (r) = 20 cm = 0.2 m - The charge of a proton (q) = 1.6 × 10^-19 C - The mass of a proton (m) = 1.67 × 10^-27 kg
02

Calculate the final kinetic energy of the proton

Using the potential difference, we can find the final kinetic energy (K) of the proton as follows: \(K = q \cdot V\) \(K = (1.6 \times 10^{-19}\,\text{C}) \times (400\,\text{V})\) \(K = 6.4 \times 10^{-17}\,\text{J}\)
03

Calculate the final velocity of the proton

Using the final kinetic energy, we can determine the final velocity (v) of the proton using the following equation: \(\frac{1}{2} mv^2 = K\) Rearranging the equation to solve for v: \(v = \sqrt{\frac{2K}{m}}\) Substituting the values for K and m: \(v = \sqrt{\frac{2 \times 6.4 \times 10^{-17}\,\text{J}}{1.67 \times 10^{-27}\,\text{kg}}}\) \(v \approx 4.38 \times 10^5\,\text{m/s}\)
04

Determine the magnetic force that causes the circular motion

The magnetic force (F) acting on the proton is equal to the centripetal force required for circular motion. We can represent this as: \(F_{\text{magnetic}} = F_{\text{centripetal}}\) \(qvB = \frac{mv^2}{r}\) Where B is the magnitude of the magnetic field, and r is the radius of the circular path.
05

Solve for the magnetic field (B)

Now, we can solve for B: \(B = \frac{mv^2}{rqv}\) Substituting the values for m, v, r, and q: \(B = \frac{(1.67 \times 10^{-27}\,\text{kg}) \times (4.38 \times 10^5\,\text{m/s})^2}{(1.6 \times 10^{-19}\,\text{C}) \times (0.2\,\text{m}) \times (4.38 \times 10^5\,\text{m/s})}\) Calculating B: \(B \approx 7.97 \times 10^{-4}\,\text{T}\) Therefore, the magnitude of the magnetic field is approximately \(7.97 \times 10^{-4}\) T.

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Most popular questions from this chapter

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?

A current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

A rail gun accelerates a projectile from rest by using the magnetic force on a current-carrying wire. The wire has radius \(r=5.1 \cdot 10^{-4} \mathrm{~m}\) and is made of copper having a density of \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\). The gun consists of rails of length \(L=1.0 \mathrm{~m}\) in a constant magnetic field of magnitude \(B=2.0 \mathrm{~T}\) oriented perpendicular to the plane defined by the rails. The wire forms an electrical connection across the rails at one end of the rails. When triggered, a current of \(1.00 \cdot 10^{4}\) A flows through the wire, which accelerates the wire along the rails. Calculate the final speed of the wire as it leaves the rails. (Neglect friction.)

An alpha particle \(\left(m=6.6 \cdot 10^{-27} \mathrm{~kg}, q=+2 e\right)\) is accelerated by a potential difference of \(2700 \mathrm{~V}\) and moves in a plane perpendicular to a constant magnetic field of magnitude \(0.340 \mathrm{~T}\), which curves the trajectory of the alpha particle. Determine the radius of curvature and the period of revolution.
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