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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

A small aluminum ball with a mass of \(5.00 \mathrm{~g}\) and a charge of \(15.0 \mathrm{C}\) is moving northward at \(3000 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.00 \mathrm{~m}\), in a clockwise sense when viewed from above. Ignoring gravity, what is the magnitude and the direction of the magnetic field that must be applied to the aluminum ball to cause it to have this motion?

A proton moving with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.40 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

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.

A proton, moving in negative \(y\) -direction in a magnetic field, experiences a force of magnitude \(F\), acting in the negative \(x\) -direction. a) What is the direction of the magnetic field producing this force? b) Does your answer change if the word "proton" in the statement is replaced by “electron"?
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