A charge Q is moving dl distance in the magnetic field B. Find the value of work done by B.
(a) 1
(b

Q: Why is the magnetic force always perpendicular to the velocity?

This is a fundamental property of the vector cross product. The magnetic force is defined as $\mathbf{F} = Q(\mathbf{v} \times \mathbf{B})$. The cross product of any two vectors always yields a third vector that is perpendicular to the plane containing the first two.

Q: If a magnetic field does no work, how do giant electromagnets pick up heavy scrap cars?

It's a great question! The magnetic field itself doesn't do the macroscopic work of lifting. The magnetic field aligns the magnetic domains in the car, creating an attractive force. However, the actual work of lifting the car against gravity is done by the crane's motor pulling the electromagnet upwards. The magnetic field just provides the 'grip'.

Q: Does the kinetic energy of the charge change as it moves in the magnetic field?

No. The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy ($W = \Delta K$). Since the work done by the magnetic field is zero, the change in kinetic energy is also zero. The particle's speed remains constant.

Q: What if the charge moves exactly parallel to the magnetic field?

If the velocity $\mathbf{v}$ is parallel to the magnetic field $\mathbf{B}$, the angle between them is $0^\circ$. The cross product $\mathbf{v} \times \mathbf{B}$ becomes zero, meaning the magnetic force is zero. If there is no force, the work done is still zero.

Q: Can a changing magnetic field do work?

A *static* magnetic field does no work. However, according to Faraday's Law of Induction, a *time-varying* magnetic field induces an electric field. It is this induced electric field that can exert a force parallel to the charge's motion and thus do work on it.

Question

A charge $Q$ is moving $dl$ distance in the magnetic field $\mathbf{B}$. Find the value of work done by $\mathbf{B}$.

Options:

(A) 1

(B) Infinite

(C) Zero

(D) -1

Sigma Percentile · Accepted Answer

### The Zero-Work Mystery of Magnetic Fields

Imagine you are observing a tiny charged particle, let's call it $Q$, zipping through a region of space filled with a magnetic field $\mathbf{B}$. As it moves, it covers an infinitesimally small distance, which we denote as $d\mathbf{l}$. A fundamental question arises: How much work does the magnetic field do on this moving charge?

To answer this, we need to dive into the mechanics of how magnetic fields interact with moving charges.

### The Lorentz Force

The force exerted by a magnetic field on a moving charge is given by the magnetic component of the Lorentz force equation:

$$\mathbf{F} = Q(\mathbf{v} 	imes \mathbf{B})$$

Here, $\mathbf{v}$ is the velocity of the charge. The crucial part of this equation is the **cross product**. In vector mathematics, the result of a cross product is a new vector that is strictly perpendicular to both of the original vectors.

Therefore, the magnetic force $\mathbf{F}$ is always perpendicular to the velocity $\mathbf{v}$, and it is also perpendicular to the magnetic field $\mathbf{B}$.

### The Geometry of Work

Now, let's recall the definition of work done in physics. Work is the dot product of the force applied and the displacement of the object:

$$dW = \mathbf{F} \cdot d\mathbf{l}$$

We know that the infinitesimal displacement $d\mathbf{l}$ is simply the velocity multiplied by an infinitesimal time interval ($d\mathbf{l} = \mathbf{v} dt$). This means that the displacement vector $d\mathbf{l}$ points in the exact same direction as the velocity vector $\mathbf{v}$.

Since we established earlier that the magnetic force $\mathbf{F}$ is perpendicular to the velocity $\mathbf{v}$, it must also be perpendicular to the displacement $d\mathbf{l}$.

Let's expand the dot product:

$$dW = |\mathbf{F}| |d\mathbf{l}| \cos	heta$$

Because the force and displacement are perpendicular, the angle $	heta$ between them is exactly $90^\circ$.

$$dW = |\mathbf{F}| |d\mathbf{l}| \cos(90^\circ)$$

Since $\cos(90^\circ) = 0$, the entire expression vanishes:

$$dW = 0$$

### The Grand Conclusion

This brings us to one of the most profound and elegant principles in electromagnetism: **Magnetic forces do no work on moving charges.**

Because the force is always pushing sideways relative to the particle's motion, it can only act as a centripetal force. It can steer the particle, forcing it to move in circles or spirals, and it can change the *direction* of the velocity. However, it can never speed the particle up or slow it down. The magnitude of the velocity (the speed) and the kinetic energy of the particle remain absolutely constant.

Animated Solution for Physics - Magnetic Effects of Current: A charge $Q$ is moving $d l$ distance in the magnetic field $B$ . Find the value of work done by $B$ .

Select Answer:

Visualized Solution

The Setup

Magnetic Force

Direction of Force

Work Done Formula

Calculating Work

Conclusion

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Zero-Work Mystery of Magnetic Fields

The Lorentz Force

The Geometry of Work

The Grand Conclusion

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Animated Solution for Physics - Magnetic Effects of Current: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A charge Q is moving dl distance in the magnetic field B. Find the value of work done by B.