Q.Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal to the field direction while the second enters at $30^\circ$ to the field direction. Name the trajectories followed by them.

Step 1 — Maximum kinetic energy (photoelectric equation)

Step 2 — Momentum of the fastest electron

Step 3 — Radius in the magnetic field …

Concept & Intuition

When a charged particle enters a uniform magnetic field at an angle, its velocity splits into two components: one parallel to the field (constant, causing uniform motion along the field) and one perpendicular (causing circular motion). The combination gives a helical path. The pitch is the distance traveled along the field in one full circular revolution.

Since both particles have the same kinetic energy and same angle, we compare how their masses and charges affect the pitch. The key is that pitch = (parallel velocity) × (time period). The time period depends on mass and charge, while the parallel velocity depends on mass (through kinetic energy). The ratio simplifies nicely.

Step-by-step reasoning

1. Write the pitch formula For a particle of mass $m$, charge $q$, entering a uniform magnetic field $B$ at angle $\theta$ to the field, the pitch $p$ is:

where $v$ is the speed, $v_{\parallel} = v \cos\theta$, and $T = \frac{2\pi m}{qB}$ is the cyclotron period.

2. Express speed in terms of kinetic energy Kinetic energy $K = \frac{1}{2} m v^2$, so $v = \sqrt{\frac{2K}{m}}$. Thus:

Since $K$ and $\theta$ are the same for both particles, and $B$ is the same field, the ratio of pitches depends only on $\frac{\sqrt{m}}{q}$.

3. Identify the particles
   • Proton: mass $m_p$, charge $+e$. …

The key idea here is that an electron moving in a circular orbit behaves like a tiny current loop. A current loop has a magnetic moment $\mu = I A$, where $I$ is the current and $A$ is the area enclosed. For a single electron orbiting with speed $v$ in a circle of radius $r$, the current is the charge passing a point per unit time: $I = e/T$, where $T = 2\pi r / v$ is the orbital period. So $I = ev/(2\pi r)$.

The area is $A = \pi r^2$, giving $\mu = (ev/(2\pi r)) \cdot \pi r^2 = \frac{evr}{2}$.

Now, the angular momentum of the electron (about the centre) is $L = mvr$. Comparing the two expressions, you see $\mu = \frac{e}{2m} L$. That ratio $\frac{e}{2m}$ is called the gyromagnetic ratio — it connects the magnetic moment to the angular momentum for any orbiting charged particle. Notice that the specific charge given is $e/m$, so $\frac{e}{2m}$ is just half of that.

Let’s work it through.

1. Write the relation The magnetic moment $\mu$ and angular momentum $L$ are linked by

This comes directly from the classical picture above — no extra assumptions needed.

2. Plug in the numbers You are given $L = 2.1 \times 10^{-34}$ Js and the specific charge $e/m = 1.76 \times 10^{11}$ C/kg. So

3. Calculate $\mu$ …

The key idea here is that current is the flow of charge, and the drift velocity of electrons is the average speed at which they move through the wire under the electric field. Once you know that speed, the time to cross the wire is just distance divided by speed.

Let’s work through it step by step.

1. Recall the relation between current and drift velocity The current $I$ in a conductor is given by

where $n$ is the number of free electrons per unit volume, $e$ is the charge of an electron ($1.6 \times 10^{-19}$ C), $A$ is the cross-sectional area, and $v_d$ is the drift velocity. This formula comes from the fact that in time $\Delta t$, the charge that crosses a section is $n e A v_d \Delta t$.

2. Solve for drift velocity Rearranging:

Plug in the values:

First compute the denominator:

$n e = (6.25 \times 10^{28})(1.6 \times 10^{-19}) = 1.0 \times 10^{10}$

Then $n e A = (1.0 \times 10^{10})(1.8 \times 10^{-6}) = 1.8 \times 10^{4}$

So

3. Find the time to travel the length of the wire The wire length $L = 120$ cm = $1.2$ m. Time $t = L / v_d$: …

The key idea here is that a single charged particle going round and round in a circle is exactly like a tiny current loop. Every time the particle completes one revolution, it has effectively transported its charge once around the circuit. So the average current is simply the charge divided by the time for one revolution — that is, the period.

Once you have that current, the magnetic field at the centre of a circular loop is a standard result from Biot-Savart law: $B = \frac{\mu_0 I}{2R}$. That’s all we need.

Let’s work it through.

1. Find the current from the moving alpha particle.

   An alpha particle has charge $q = +2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$ C.

   The time period $T = 3.14\ \mu\text{s} = 3.14 \times 10^{-6}$ s.

   Current $I = \frac{q}{T} = \frac{3.2 \times 10^{-19}}{3.14 \times 10^{-6}}$.

   Notice $3.14 \approx \pi$, so $I = \frac{3.2 \times 10^{-19}}{\pi \times 10^{-6}} = \frac{3.2}{\pi} \times 10^{-13}$ A.

2. Apply the formula for the magnetic field at the centre of a circular loop.

   Radius $R = 2\ \text{fermi} = 2 \times 10^{-15}$ m.

   $B = \frac{\mu_0 I}{2R}$, with $\mu_0 = 4\pi \times 10^{-7}$ T m/A.

   Substitute:

3. Simplify step by step. The $\pi$ cancels: $4\pi \times \frac{3.2}{\pi} = 4 \times 3.2 = 12.8$. So numerator becomes $12.8 \times 10^{-7} \times 10^{-13} = 12.8 \times 10^{-20}$. …

The key idea here is that a charged particle in a uniform electric field experiences a constant force, just like a mass in a uniform gravitational field. When the velocity is opposite to the field direction, the force acts as a brake — a constant deceleration. This turns the problem into a straightforward one-dimensional kinematics or work-energy problem.

Let’s unpack it step by step.

1. Identify the force and acceleration The electric force on a charge $q$ in a uniform field $E$ is $F = qE$. Since the charge is positive ($20\,\mu\text{C}$) and the field is uniform, the force is in the direction of the field. But the particle is thrown opposite to the field, so the force opposes the motion — it decelerates the particle. The magnitude of the force:

Mass $m = 5\,\text{g} = 5 \times 10^{-3}\,\text{kg}$.

Deceleration (magnitude):

2. Choose the right equation of motion Initial velocity $u = 16\,\text{m/s}$ (in the direction we call positive). Final velocity $v = 0$. Acceleration is negative relative to the direction of motion: $a = -800\,\text{m/s}^2$. Using $v^2 = u^2 + 2as$:

$1600s = 256$ …

Concept & Intuition

This problem combines two classic ideas:

• Photoelectric effect: A photon’s energy ($h\nu$) is used to overcome the work function ($\phi$) and give the electron kinetic energy ($K_{\text{max}}$).
• Circular motion in a magnetic field: The maximum kinetic energy corresponds to the largest possible radius of the electron’s path, because $r = \frac{mv}{qB}$.

The key link: the electron’s speed $v$ from the photoelectric effect determines the radius $r$ in the magnetic field. By measuring $r$, we can find $K_{\text{max}}$, and then $\phi = h\nu - K_{\text{max}}$.

Step-by-step solution

1. Find the photon energy Wavelength $\lambda = 2480 \, \text{Å} = 2480 \times 10^{-10} \, \text{m} = 2.48 \times 10^{-7} \, \text{m}$. Photon energy:

Using $hc = 12400 \, \text{eV·Å}$ (a handy constant):

2. Relate the magnetic radius to kinetic energy For a charged particle in a perpendicular magnetic field:

Here $q = e$, $m = m_e$. The kinetic energy is

Eliminate $v$: from $v = \frac{eBr}{m}$,

3. Plug in the numbers Given: $B = \frac{1}{4} \times 10^{-5} \, \text{T} = 2.5 \times 10^{-6} \, \text{T}$, $r = 1 \, \text{m}$. Use $e = 1.6 \times 10^{-19} \, \text{C}$, $m_e = 9.1 \times 10^{-31} \, \text{kg}$.

Compute stepwise:

• $e^2 = 2.56 \times 10^{-38}$
• $B^2 = 6.25 \times 10^{-12}$
• Numerator: $2.56 \times 10^{-38} \times 6.25 \times 10^{-12} = 1.6 \times 10^{-49}$ …

The problem asks for the induced emf in a cycle wheel rim rolling along a horizontal road. The rim is a closed circular loop. When a closed loop moves in a uniform magnetic field, the induced emf depends on the change of magnetic flux through the loop. Here, the Earth’s magnetic field is uniform over the size of the wheel, and the wheel’s motion is purely translational (the rolling does not change the orientation of the loop relative to the field). Therefore, the flux through the rim remains constant, and no net emf is induced.

Let’s work through the reasoning step by step.

1. Understand the setup

   The rim is a circular loop of radius $r = 40\ \text{cm} = 0.4\ \text{m}$. It rolls north with angular speed $\omega = 20\ \text{rad/s}$. The horizontal component of Earth’s magnetic field is $B_H = 0.26\ \text{G} = 0.26 \times 10^{-4}\ \text{T} = 2.6 \times 10^{-5}\ \text{T}$ (since $1\ \text{G} = 10^{-4}\ \text{T}$). The field is horizontal and directed toward north (magnetic north). The wheel rolls north, so its velocity is parallel to the horizontal field.

2. Recall the condition for induced emf in a moving loop

   Faraday’s law states that the induced emf in a closed loop is $\mathcal{E} = -\frac{d\Phi}{dt}$, where $\Phi$ is the magnetic flux through the loop. For a loop moving in a uniform magnetic field, the flux changes only if the area perpendicular to the field changes or if the loop rotates relative to the field. Here, the loop is circular and remains in a vertical plane; its orientation relative to the horizontal field does not change as it rolls. The area vector of the loop is horizontal (perpendicular to the plane of the wheel), but the field is also horizontal. So the flux is $\Phi = B_H \cdot A \cdot \cos\theta$, where $\theta$ is the angle between the field direction and the area vector. Since both are horizontal and the wheel rolls north, the area vector points east-west (depending on which side you take), but the field is north-south. Thus $\theta = 90^\circ$ and $\cos 90^\circ = 0$, so the flux is zero initially and remains zero throughout the motion.

3. Consider the possibility of motional emf in parts of the rim

   One might think that each small segment of the rim moving through the magnetic field develops a motional emf $\mathcal{E} = (\vec{v} \times \vec{B}) \cdot d\vec{l}$. However, for a closed loop, these contributions cancel out if the flux is constant. In this case, because the entire loop moves together without changing shape or orientation, the net emf around the loop is zero. This is analogous to a wire loop moving uniformly in a uniform field — no net emf is induced. …