101. A charged particle enters crossed fields with \(\vec{v}\perp\vec{B}\). The magnitudes are \(E=8.0\times10^4\,\text{N C}^{-1}\) and \(B=0.40\,\text{T}\). If its speed is \(1.5\times10^5\,\text{m s}^{-1}\), then compared with the magnetic force, the electric force is
ⓐ. smaller
ⓑ. equal
ⓒ. larger
ⓓ. zero
Correct Answer: larger
Explanation: \( \textbf{Given data:} \) \(E=8.0\times10^4\,\text{N C}^{-1}\), \(B=0.40\,\text{T}\), and \(v=1.5\times10^5\,\text{m s}^{-1}\).
\( \textbf{Selected-speed relation:} \)
\[
v_0=\frac{E}{B}
\]
\( \textbf{Calculate selected speed:} \)
\[
v_0=\frac{8.0\times10^4}{0.40}=2.0\times10^5\,\text{m s}^{-1}
\]
\( \textbf{Compare speeds:} \)
\[
v=1.5\times10^5\,\text{m s}^{-1}\lt v_0
\]
\( \textbf{Force comparison:} \) At \(v_0\), \(qE=qv_0B\).
For a smaller speed, \(qvB\) is smaller than \(qE\).
Therefore the electric force magnitude is larger than the magnetic force magnitude.
The charge value is not needed because it would multiply both force magnitudes.
\( \textbf{Final answer:} \) The electric force is larger.
102. A velocity selector has \(E=1.2\times10^5\,\text{N C}^{-1}\). Particles of speed \(3.0\times10^5\,\text{m s}^{-1}\) must pass undeflected. The required magnetic field magnitude is
ⓐ. \(0.40\,\text{T}\)
ⓑ. \(0.20\,\text{T}\)
ⓒ. \(0.60\,\text{T}\)
ⓓ. \(2.50\,\text{T}\)
Correct Answer: \(0.40\,\text{T}\)
Explanation: \( \textbf{Given data:} \) \(E=1.2\times10^5\,\text{N C}^{-1}\) and selected speed \(v=3.0\times10^5\,\text{m s}^{-1}\).
\( \textbf{Condition for undeflected motion:} \)
\[
qE=qvB
\]
\( \textbf{Cancel charge:} \)
\[
E=vB
\]
\( \textbf{Solve for magnetic field:} \)
\[
B=\frac{E}{v}
\]
\( \textbf{Substitution:} \)
\[
B=\frac{1.2\times10^5}{3.0\times10^5}
\]
\( \textbf{Simplification:} \)
\[
B=\frac{1.2}{3.0}=0.40
\]
\[
B=0.40\,\text{T}
\]
The unit is tesla because \(E/v\) has the same unit as magnetic field in the selector relation.
\( \textbf{Final answer:} \) The required magnetic field is \(0.40\,\text{T}\).
103. Read the following passage.
A beam contains charged particles of different speeds. It enters a region where uniform \(\vec{E}\) and \(\vec{B}\) are perpendicular to each other and also arranged so that electric and magnetic forces oppose for the chosen path.
Particles emerging undeflected from the region have
ⓐ. the same mass
ⓑ. speed \(\frac{E}{B}\)
ⓒ. the same charge magnitude
ⓓ. the same kinetic energy
Correct Answer: speed \(\frac{E}{B}\)
Explanation: The condition for undeflected motion is equality of electric and magnetic force magnitudes in opposite directions. This gives \(qE=qvB\). Cancelling \(q\) leads to \(v=\frac{E}{B}\), so the selector chooses speed. It does not require all undeflected particles to have the same mass or the same charge magnitude. Kinetic energy can differ for particles of different masses even if their selected speed is the same.
104. A cyclotron accelerates charged particles using
ⓐ. a steady magnetic field in the gap and electric field in each dee
ⓑ. an electric field across the gap and a magnetic field in the dees
ⓒ. a gravitational field in the gap and magnetic field outside the dees
ⓓ. a magnetic field doing work inside each dee and no gap field
Correct Answer: an electric field across the gap and a magnetic field in the dees
Explanation: A cyclotron uses two main field roles. The magnetic field is perpendicular to the plane of the dees and bends the charged particle into circular or semicircular paths inside the dees. The electric field acts mainly across the gap between the dees and accelerates the particle when it crosses the gap. The electric field must alternate so that it accelerates the particle each time it reaches the gap. The magnetic field guides the path, while the electric field supplies energy.
105. Inside the dees of a cyclotron, the charged particle usually moves in a circular arc because
ⓐ. the magnetic field supplies the centripetal force
ⓑ. the electric field inside each dee continuously increases speed
ⓒ. the particle becomes neutral inside each dee
ⓓ. the magnetic field does positive work on the particle
Correct Answer: the magnetic field supplies the centripetal force
Explanation: In the dees of a cyclotron, the magnetic field is perpendicular to the particle’s velocity. This magnetic force is perpendicular to the velocity and acts as the centripetal force required for circular motion. Since magnetic force does no work, it does not directly increase the particle’s kinetic energy inside the dee. The particle gains energy mainly when it crosses the gap between the dees due to the electric field. The dee region bends the path, while the gap region accelerates the particle.
106. The resonance condition in a cyclotron requires the frequency of the alternating electric field to match the particle’s
ⓐ. speed at the outermost orbit
ⓑ. kinetic energy at the centre
ⓒ. radius in the first semicircle
ⓓ. cyclotron frequency
Correct Answer: cyclotron frequency
Explanation: The charged particle returns to the gap after each half revolution. For repeated acceleration, the electric field must reverse at the right time so that it pushes the particle forward whenever it crosses the gap. This timing is achieved when the oscillator frequency matches the cyclotron frequency. The cyclotron frequency is \(f=\frac{|q|B}{2\pi m}\) in the non-relativistic case. Matching frequency is about timing, not about matching radius or kinetic energy directly.
107. For a non-relativistic charged particle in a cyclotron, the cyclotron frequency is
ⓐ. \(f=\frac{2\pi m}{|q|B}\)
ⓑ. \(f=\frac{|q|B}{2\pi m}\)
ⓒ. \(f=\frac{mv}{|q|BR}\)
ⓓ. \(f=\frac{|q|BR}{m}\)
Correct Answer: \(f=\frac{|q|B}{2\pi m}\)
Explanation: In a uniform magnetic field, the angular frequency of circular motion is \(\omega=\frac{|q|B}{m}\). Frequency and angular frequency are related by \(\omega=2\pi f\). Therefore, \(f=\frac{\omega}{2\pi}=\frac{|q|B}{2\pi m}\). The expression contains charge magnitude, magnetic field, and mass, but not speed or radius. This speed independence is what allows a fixed oscillator frequency to accelerate the particle repeatedly in the non-relativistic limit.
108. Assertion: In an ideal non-relativistic cyclotron, the time period of revolution is independent of the particle’s speed.
Reason: As the speed increases, the radius increases in the same proportion.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The time period of circular motion in a uniform magnetic field is \(T=\frac{2\pi m}{|q|B}\). This expression does not contain speed, so the Assertion is true in the non-relativistic case. The radius is \(r=\frac{mv}{|q|B}\), so increasing speed increases radius proportionally. The circumference of the circular path increases in proportion to \(v\), while the particle also moves faster by the same factor. That proportional increase explains why the time for one revolution remains unchanged.
109. Study the table about field roles in a cyclotron.
| Row | Part of cyclotron | Main role |
|---|
| P | Magnetic field in the dees | Bends the path into circular arcs |
| Q | Electric field across the gap | Accelerates the charged particle |
| R | Alternating oscillator | Maintains proper timing of acceleration |
| S | Magnetic field inside the dees | Directly increases kinetic energy by doing work |
The row that gives an incorrect role is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: The magnetic field in the dees bends the charged particle path because the magnetic force acts as centripetal force. The electric field across the gap accelerates the particle and increases its kinetic energy. The alternating oscillator is needed so that the electric field has the proper direction whenever the particle crosses the gap. Row S is wrong because magnetic force is perpendicular to velocity and does no work. The energy gain in a cyclotron is due to the electric field, not the magnetic field.
110. A charged particle of mass \(3.2\times10^{-27}\,\text{kg}\) and charge magnitude \(1.6\times10^{-19}\,\text{C}\) moves in a cyclotron with \(B=1.0\,\text{T}\). Using \(\pi\approx3.14\), the cyclotron frequency is closest to
ⓐ. \(4.0\times10^6\,\text{Hz}\)
ⓑ. \(8.0\times10^6\,\text{Hz}\)
ⓒ. \(1.6\times10^7\,\text{Hz}\)
ⓓ. \(3.2\times10^7\,\text{Hz}\)
Correct Answer: \(8.0\times10^6\,\text{Hz}\)
Explanation: \( \textbf{Given data:} \) \(|q|=1.6\times10^{-19}\,\text{C}\), \(m=3.2\times10^{-27}\,\text{kg}\), \(B=1.0\,\text{T}\), and \(\pi\approx3.14\).
\( \textbf{Required:} \) Cyclotron frequency \(f\).
\( \textbf{Relation:} \)
\[
f=\frac{|q|B}{2\pi m}
\]
\( \textbf{Substitution:} \)
\[
f=\frac{(1.6\times10^{-19})(1.0)}{(2)(3.14)(3.2\times10^{-27})}
\]
\( \textbf{Denominator:} \)
\[
(2)(3.14)(3.2)\times10^{-27}=20.096\times10^{-27}
\]
\[
20.096\times10^{-27}=2.0096\times10^{-26}
\]
\( \textbf{Division:} \)
\[
f=\frac{1.6\times10^{-19}}{2.0096\times10^{-26}}\approx0.796\times10^7
\]
\[
f\approx7.96\times10^6\,\text{Hz}
\]
The closest option is \(8.0\times10^6\,\text{Hz}\).
\( \textbf{Final answer:} \) The cyclotron frequency is closest to \(8.0\times10^6\,\text{Hz}\).
111. If the magnetic field of a cyclotron is doubled while the particle species remains the same, the required oscillator frequency should
ⓐ. become half
ⓑ. remain unchanged
ⓒ. become double
ⓓ. become four times
Correct Answer: become double
Explanation: The cyclotron frequency is \(f=\frac{|q|B}{2\pi m}\). For the same particle species, \(|q|\) and \(m\) are fixed. Therefore, \(f\propto B\). If the magnetic field is doubled, the cyclotron frequency also doubles. The oscillator frequency must follow this change to keep the electric field reversal synchronized with the particle’s motion.
112. A cyclotron is designed for protons. If an alpha particle is used in the same magnetic field, its cyclotron frequency compared with that of the proton is approximately
ⓐ. half the proton frequency
ⓑ. twice the proton frequency
ⓒ. equal to the proton frequency
ⓓ. four times the proton frequency
Correct Answer: half the proton frequency
Explanation: Cyclotron frequency is \(f=\frac{|q|B}{2\pi m}\). A proton has charge magnitude \(e\) and mass approximately \(m_p\). An alpha particle has charge magnitude \(2e\) and mass approximately \(4m_p\). Therefore, its \(\frac{|q|}{m}\) ratio is \(\frac{2e}{4m_p}=\frac{1}{2}\frac{e}{m_p}\). In the same magnetic field, the alpha-particle frequency is about half the proton frequency.
113. A charged particle completes one semicircle inside a dee of a cyclotron. The time taken for this semicircle is
ⓐ. \(t=\frac{2\pi m}{|q|B}\)
ⓑ. \(t=\frac{|q|B}{\pi m}\)
ⓒ. \(t=\frac{\pi m}{|q|B}\)
ⓓ. \(t=\frac{m}{2\pi |q|B}\)
Correct Answer: \(t=\frac{\pi m}{|q|B}\)
Explanation: The time period for one full circular revolution in the magnetic field is \(T=\frac{2\pi m}{|q|B}\). A semicircle is half of a full revolution. Therefore, the time taken for one semicircle is \(\frac{T}{2}\). Substituting the expression for \(T\) gives \(\frac{T}{2}=\frac{\pi m}{|q|B}\). This time determines when the particle returns to the gap and why the alternating electric field must reverse at the proper rate.
114. In the non-relativistic operation of a cyclotron, the particle moves in spirals of increasing radius. The radius increases mainly because
ⓐ. magnetic field doing work inside the dee
ⓑ. charge magnitude increasing after each turn
ⓒ. electric field at the gap increasing speed
ⓓ. mass decreasing after each crossing
Correct Answer: electric field at the gap increasing speed
Explanation: The radius of circular motion in the magnetic field is \(r=\frac{mv}{|q|B}\). For the same particle in the same magnetic field, radius is directly proportional to speed. The particle gains speed when it is accelerated by the electric field across the gap. After each acceleration, it enters the dee with a larger speed and therefore follows a larger-radius semicircle. The spiral growth is caused by energy gain from the electric field, while the magnetic field bends the path.
115. In a cyclotron, if the final orbit radius is \(R\), the maximum speed of a non-relativistic particle is
ⓐ. \(v_{\max}=\frac{|q|BR}{m}\)
ⓑ. \(v_{\max}=\frac{mBR}{|q|}\)
ⓒ. \(v_{\max}=\frac{|q|B}{mR}\)
ⓓ. \(v_{\max}=\frac{mR}{|q|B}\)
Correct Answer: \(v_{\max}=\frac{|q|BR}{m}\)
Explanation: The radius of circular motion in a uniform magnetic field is \(r=\frac{mv}{|q|B}\). In a cyclotron, the largest possible speed is reached near the largest available orbit radius \(R\). Replacing \(r\) by \(R\) gives \(R=\frac{mv_{\max}}{|q|B}\). Solving this relation gives \(v_{\max}=\frac{|q|BR}{m}\). The expression shows that a larger magnetic field or larger dee radius allows a greater final speed for the same particle.
116. The maximum kinetic energy of a non-relativistic charged particle emerging from a cyclotron of radius \(R\) is
ⓐ. \(K_{\max}=\frac{|q|BR}{m}\)
ⓑ. \(K_{\max}=\frac{2m}{q^2B^2R^2}\)
ⓒ. \(K_{\max}=\frac{|q|B^2R}{2m}\)
ⓓ. \(K_{\max}=\frac{q^2B^2R^2}{2m}\)
Correct Answer: \(K_{\max}=\frac{q^2B^2R^2}{2m}\)
Explanation: The maximum speed in a cyclotron is \(v_{\max}=\frac{|q|BR}{m}\). The kinetic energy is \(K=\frac{1}{2}mv^2\). Substituting the maximum speed gives \(K_{\max}=\frac{1}{2}m\left(\frac{|q|BR}{m}\right)^2\). This simplifies to \(K_{\max}=\frac{q^2B^2R^2}{2m}\), since the square of charge magnitude is \(q^2\). The squared dependence on \(B\) and \(R\) means doubling either one increases \(K_{\max}\) by a factor of \(4\).
117. Study the table about maximum energy in a cyclotron.
| Row | Change made | Effect on \(K_{\max}\) |
|---|
| P | \(B\) doubled | \(K_{\max}\) becomes \(4\) times |
| Q | \(R\) doubled | \(K_{\max}\) becomes \(4\) times |
| R | \(|q|\) doubled with \(m\) unchanged | \(K_{\max}\) becomes \(4\) times |
| S | \(m\) doubled with \(|q|\), \(B\), and \(R\) unchanged | \(K_{\max}\) becomes \(2\) times |
The row with the incorrect dependence is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row S
ⓓ. Row R
Correct Answer: Row S
Explanation: The maximum kinetic energy is \(K_{\max}=\frac{q^2B^2R^2}{2m}\). Since \(K_{\max}\propto B^2\), doubling \(B\) makes the energy \(4\) times, so row P is valid. Since \(K_{\max}\propto R^2\), doubling \(R\) also makes the energy \(4\) times, so row Q is valid. Since \(K_{\max}\propto q^2\), doubling \(|q|\) makes the energy \(4\) times if mass is unchanged. Row S is wrong because \(K_{\max}\propto\frac{1}{m}\), so doubling \(m\) would make the maximum energy half, not double.
118. A cyclotron has magnetic field \(B=0.50\,\text{T}\) and final orbit radius \(R=0.80\,\text{m}\). A particle with \(|q|=1.6\times10^{-19}\,\text{C}\) and \(m=3.2\times10^{-27}\,\text{kg}\) reaches the outer edge. Its maximum speed is
ⓐ. \(1.0\times10^7\,\text{m s}^{-1}\)
ⓑ. \(4.0\times10^7\,\text{m s}^{-1}\)
ⓒ. \(8.0\times10^7\,\text{m s}^{-1}\)
ⓓ. \(2.0\times10^7\,\text{m s}^{-1}\)
Correct Answer: \(2.0\times10^7\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(|q|=1.6\times10^{-19}\,\text{C}\), \(B=0.50\,\text{T}\), \(R=0.80\,\text{m}\), and \(m=3.2\times10^{-27}\,\text{kg}\).
\( \textbf{Required:} \) Maximum speed \(v_{\max}\).
\( \textbf{Relation at outer radius:} \)
\[
v_{\max}=\frac{|q|BR}{m}
\]
\( \textbf{Substitution:} \)
\[
v_{\max}=\frac{(1.6\times10^{-19})(0.50)(0.80)}{3.2\times10^{-27}}
\]
\( \textbf{Numerator:} \)
\[
(1.6)(0.50)(0.80)=0.64
\]
\[
0.64\times10^{-19}=6.4\times10^{-20}
\]
\( \textbf{Division:} \)
\[
v_{\max}=\frac{6.4\times10^{-20}}{3.2\times10^{-27}}=2.0\times10^7\,\text{m s}^{-1}
\]
The final radius is used because the largest speed occurs at the largest circular path inside the cyclotron.
\( \textbf{Final answer:} \) The maximum speed is \(2.0\times10^7\,\text{m s}^{-1}\).
119. A charged particle in a cyclotron has \(K_{\max}=K\). If the cyclotron radius is doubled while \(B\), \(|q|\), and \(m\) remain unchanged, the new maximum kinetic energy is
ⓐ. \(\frac{K}{4}\)
ⓑ. \(\frac{K}{2}\)
ⓒ. \(2K\)
ⓓ. \(4K\)
Correct Answer: \(4K\)
Explanation: \( \textbf{Maximum energy relation:} \)
\[
K_{\max}=\frac{q^2B^2R^2}{2m}
\]
\( \textbf{Fixed quantities:} \) \(|q|\), \(B\), and \(m\) remain unchanged.
\( \textbf{Dependence on radius:} \)
\[
K_{\max}\propto R^2
\]
\( \textbf{Radius change:} \) \(R\) changes to \(2R\).
\( \textbf{Energy factor:} \)
\[
\frac{K'_{\max}}{K_{\max}}=\left(\frac{2R}{R}\right)^2=4
\]
So the new maximum kinetic energy is \(4K\).
The square appears because speed is proportional to \(R\), while kinetic energy is proportional to \(v^2\).
\( \textbf{Final answer:} \) The new maximum kinetic energy is \(4K\).
120. A cyclotron operating at high particle speeds becomes less effective when relativistic effects become significant because
ⓐ. magnetic field becomes unable to keep circular motion
ⓑ. relativistic inertia changes, disturbing resonance
ⓒ. electric field cannot reverse at any frequency
ⓓ. particle charge becomes zero at high speed
Correct Answer: relativistic inertia changes, disturbing resonance
Explanation: The ordinary cyclotron frequency is \(f=\frac{|q|B}{2\pi m}\), which assumes a constant particle mass. At sufficiently high speeds, relativistic effects make the effective inertia increase. This changes the actual revolution frequency, so a fixed oscillator frequency no longer stays perfectly synchronized with the particle’s arrival at the gap. The magnetic field still bends the path, and the electric field can still exist, but the timing condition is disturbed. This is why the simple cyclotron formula is treated as a non-relativistic result.
