An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kin

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Q. An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

(a) the two magnetic forces are equal and opposite, so they produce no net effect.

(b) the magnetic forces do no work on each particle.

(d) the magenetic forces are necessarily negligible.

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't)

It tells you: How much the speed changes when you know the net work done. Or, how much net work is needed to achieve a certain speed change.

It doesn't tell you: The direction of motion, the time taken, or the path followed. Work and kinetic energy are scalars — they have no direction.

A Quick Example

A 2 kg block initially at rest is pulled by a net force of 10 N over 4 m. Find its final speed.

Solution:

Net work: $W = Fs = 10 \times 4 = 40 \text{ J}$
Initial kinetic energy: $K_i = 0$
By the theorem: $40 = \frac{1}{2}(2)v_f^2 - 0$
So: $40 = v_f^2$
Therefore: $v_f = \sqrt{40} \approx 6.32 \text{ m/s}$

Important

The Work-Energy Theorem is a scalar alternative to Newton's laws for problems involving speed changes. It often simplifies calculations because you don't need to find acceleration or time — just work and kinetic energy.

Looking up "Work Energy Theorem: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Work Energy Theorem is drawn directly from the Work, Energy and Power coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

Magnetic forces are always perpendicular to velocity, so they do zero work on any charged particle. The change in kinetic energy depends only on work done, making the magnetic interaction irrelevant to energy calculations. The answer is (B).

Why magnetic forces don't change kinetic energy

The work-energy theorem tells us that the change in kinetic energy of a particle equals the net work done on it:

\Delta KE = W_{\text{net}} = \int \vec{F} \cdot d\vec{s}

Work is the dot product of force and displacement. When a force acts perpendicular to motion, that dot product vanishes—the force does nothing to speed up or slow down the particle.

The magnetic force on a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force:

\vec{F}_{\text{mag}} = q(\vec{v} \times \vec{B})

The cross product $\vec{v} \times \vec{B}$ produces a vector perpendicular to both $\vec{v}$ and $\vec{B}$ . This means the magnetic force is always perpendicular to the particle's instantaneous velocity.

Step-by-step reasoning

Calculate the work done by the magnetic force on one particle. The instantaneous power delivered by any force is $P = \vec{F} \cdot \vec{v}$ . For the magnetic force:

P_{\text{mag}} = \vec{F}_{\text{mag}} \cdot \vec{v} = q(\vec{v} \times \vec{B}) \cdot \vec{v}

The scalar triple product $(\vec{v} \times \vec{B}) \cdot \vec{v} = 0$ because $\vec{v} \times \vec{B}$ is perpendicular to $\vec{v}$ . Therefore $P_{\text{mag}} = 0$ at every instant.

Integrate over the path. Since the instantaneous power is zero everywhere along the trajectory, the total work done is:

W_{\text{mag}} = \int P_{\text{mag}} \, dt = 0

Apply to both particles.

The proton creates a magnetic field that exerts a force on the electron, and vice versa. But each magnetic force is perpendicular to the velocity of the particle it acts upon. Both particles experience zero work from the magnetic interaction.
Kinetic energy of the system.

Since neither particle gains or loses kinetic energy from the magnetic forces, these forces do not contribute to $\Delta KE_{\text{system}}$ . Only the electric (Coulomb) forces, which can do work, change the kinetic energy.

Watch out

Option (A) might seem tempting because Newton's third law guarantees the magnetic forces are equal and opposite. But equal-and-opposite forces can still do net work on a system if the particles move different distances or in different directions (think of a compressed spring pushing two blocks apart). The real reason is that each force individually does zero work.

Tip

Magnetic forces can change the direction of motion (they provide centripetal acceleration in circular paths) but never the speed. They deflect without energizing.

✓Final answer

The correct option is (B) : the magnetic forces do no work on each particle.

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Q.An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

The Intuition First

The Precise Statement

Why "Net" Work Matters

A Simple Derivation (for constant force)

What It Tells You (and What It Doesn't)

A Quick Example

Why magnetic forces don't change kinetic energy

Step-by-step reasoning

Concept: Work Done by a Magnetic Force is Always Zero

Q.An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

Why magnetic forces don't change kinetic energy

Step-by-step reasoning

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