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Chapter 19: Problem 27

A singly charged ion of unknown mass moves in a circle of radius $12.5 \mathrm{cm}\( in a magnetic field of \)1.2 \mathrm{T}$. The ion was accelerated through a potential difference of \(7.0 \mathrm{kV}\) before it entered the magnetic field. What is the mass of the ion?

Short Answer

Expert verified
Answer: The mass of the ion is approximately \(9.24 × 10^{-27}\ kg\).

Step by step solution

01

Write down the formula for magnetic force

The magnetic force acting on a charged particle in a magnetic field is given by the formula: \[F_m = qvB\sin(\theta)\] where \(F_m\) = magnetic force (N) \(q\) = charge of the particle (C) \(v\) = velocity of the particle (m/s) \(B\) = magnetic field (T) \(\theta\) = angle between the velocity vector and magnetic field vector In this case, the ion is singly charged, so its charge is equal to the elementary charge (\(q = 1.6 × 10^{-19} C\)). The ion moves in a circular path, so the angle between the velocity vector and the magnetic field vector is \(90^\circ\). Therefore, \(\sin(\theta) = 1\).
02

Write down the formula for centripetal force

The centripetal force required for the particle's circular motion is given by the formula: \[F_c = \frac{mv^2}{r}\] where \(F_c\) = centripetal force (N) \(m\) = mass of the particle (kg) \(v\) = velocity of the particle (m/s) \(r\) = radius of the circular path (m)
03

Equate magnetic force and centripetal force

Since the magnetic force is responsible for the centripetal force, we can equate the two forces: \[qvB = \frac{mv^2}{r}\]
04

Solve for velocity

Before we can find the mass, we need to determine the velocity of the ion, which can be calculated using the potential difference the ion was accelerated through: \[V = \frac{1}{2}mv^2\] Solve for \(v\): \[v = \sqrt{\frac{2qV}{m}}\]
05

Substitute the velocity expression into the magnetic force equation

Now substitute the velocity expression from Step 4 into the equation from Step 3: \[qB = \frac{m\left(\sqrt{\frac{2qV}{m}}\right)^2}{r}\]
06

Solve for mass

Now we can solve for the mass: \[m = \frac{2q^2Vr}{B^2}\]
07

Substitute the given values into the mass equation and find the mass

Substitute the given values into the mass equation: \(q = 1.6 × 10^{-19}\ C\) \(V = 7.0 × 10^3\ V\) \(B = 1.2\ T\) \(r = 0.125\ m\) \[m = \frac{2(1.6 × 10^{-19}\ C)^2(7.0 × 10^3\ V)(0.125\ m)}{(1.2\ T)^2}\] After calculating, we get: \[m \approx 9.24 × 10^{-27} kg\] The mass of the ion is approximately \(9.24 × 10^{-27}\ kg\).

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Most popular questions from this chapter

A strip of copper \(2.0 \mathrm{cm}\) wide carries a current \(I=\) $30.0 \mathrm{A}\( to the right. The strip is in a magnetic field \)B=5.0 \mathrm{T}$ into the page. (a) What is the direction of the average magnetic force on the conduction electrons? (b) The Hall voltage is \(20.0 \mu \mathrm{V}\). What is the drift velocity?
An electromagnetic flowmeter is to be used to measure blood speed. A magnetic field of \(0.115 \mathrm{T}\) is applied across an artery of inner diameter \(3.80 \mathrm{mm}\) The Hall voltage is measured to be \(88.0 \mu \mathrm{V} .\) What is the average speed of the blood flowing in the artery?
A charged particle is accelerated from rest through a potential difference \(\Delta V\). The particle then passes straight through a velocity selector (field magnitudes \(E\) and \(B\) ). Derive an expression for the charge-to-mass ratio \((q / m)\) of the particle in terms of \(\Delta V, E,\) and \(B\)
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish $^{12} \mathrm{C}^{+}\( and \)^{14} \mathrm{C}^{+}$ ions that have the same charge. The \(^{12} \mathrm{C}^{+}\) ions move in a circle of diameter \(25 \mathrm{cm} .\) (a) What is the diameter of the orbit of \(^{14} \mathrm{C}^{+}\) ions? (b) What is the ratio of the frequencies of revolution for the two types of ion?
Point \(P\) is midway between two long, straight, parallel wires that run north- south in a horizontal plane. The distance between the wires is $1.0 \mathrm{cm} .\( Each wire carrics a current of \)1.0 \mathrm{A}$ toward the north. Find the magnitude and direction of the magnetic field at point \(P\)
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