A conductor lies along the z-axis at -1.5

Q: Why is the length vector taken as negative?

Because the current is flowing in the $-a_z$ direction, and the length vector must point in the direction of the current flow to correctly apply the $\mathbf{L} \times \mathbf{B}$ formula.

Q: Why do we integrate to find the work done?

The magnetic field $\mathbf{B}$ depends on $x$ ($e^{-0.2x}$), which means the magnetic force changes as the conductor moves along the x-axis. For a variable force, work done is strictly the integral of force with respect to displacement.

Q: Does the magnetic field do positive or negative work?

The magnetic force is in the $+x$ direction, and the displacement is also in the $+x$ direction. Therefore, the magnetic field does positive work on the conductor.

Q: What does 'power required to move the conductor' mean here?

It refers to the rate at which energy is being transferred in the system. Since the magnetic force is doing positive work, an external agent would actually need to extract power to keep the speed constant. The question asks for the magnitude of this power transfer.

Q: How do we calculate $e^{-0.4}$ without a calculator in an exam?

In JEE exams, such values are either provided in the question, or you can use the Taylor series expansion $e^x \approx 1 + x + x^2/2!$ for small $x$. Here, $e^{-0.4} \approx 1 - 0.4 + 0.16/2 = 0.68$, which gives $P = 9(1 - 0.68) = 2.88$ W, a close enough approximation to confidently select the $2.97$ W option.

Question

A conductor lies along the z-axis at $-1.5 \le z < 1.5$ m and carries a fixed current of $10.0$ A in $-a_z$ direction (see figure). For a field $B = 3.0 	imes 10^{-4} e^{-0.2x} a_y$ T, find the power required to move the conductor at constant speed to $x = 2.0$ m, $y = 0$ in $5 	imes 10^{-3}$ s. Assume parallel motion along the x-axis.

Options:

(A) 1.57 W

(B) 2.97 W

(C) 14.85 W

(D) 29.7 W

Sigma Percentile · Accepted Answer

Imagine you are an engineer tasked with moving a massive conductor through a magnetic field. This isn't just any magnetic field; its strength fades away exponentially as you push forward. How much power would you need to maintain a constant speed? Let's break down the physics and mathematics behind this fascinating problem.

### Visualizing the 3D Setup

First, let's orient ourselves in 3D space. We have a straight conductor lying along the z-axis, stretching from $z = -1.5$ m to $z = 1.5$ m. This gives it a total length of $L = 3.0$ m. A steady current of $I = 10.0$ A flows through it in the negative z-direction, so we can define our length vector as $\mathbf{L} = -3.0 \hat{k}$ m.

The entire region is permeated by a magnetic field pointing in the positive y-direction. However, its magnitude is not constant; it decays exponentially along the x-axis:

$$\mathbf{B} = 3.0 	imes 10^{-4} e^{-0.2x} \hat{j} 	ext{ T}$$

### The Lorentz Force in Action

When a current-carrying wire sits in a magnetic field, it experiences a magnetic force (Lorentz force) given by the cross product:

$$\mathbf{F}_m = I (\mathbf{L} 	imes \mathbf{B})$$

Let's substitute our known vectors into this equation:

$$\mathbf{F}_m = 10 	imes (-3.0 \hat{k}) 	imes (3.0 	imes 10^{-4} e^{-0.2x} \hat{j})$$

Grouping the scalar terms together, we get:

$$\mathbf{F}_m = -90 	imes 10^{-4} e^{-0.2x} (\hat{k} 	imes \hat{j})$$

Using the right-hand rule for cross products, we know that $\hat{k} 	imes \hat{j} = -\hat{i}$. The two negative signs cancel out beautifully, leaving us with a force pointing purely in the positive x-direction:

$$\mathbf{F}_m = 9.0 	imes 10^{-3} e^{-0.2x} \hat{i} 	ext{ N}$$

### The Calculus of Work

We need to move this conductor from $x = 0$ to $x = 2.0$ m at a constant speed. Because the magnetic force is pushing the conductor in the $+x$ direction, an external agent would actually need to pull back to prevent it from accelerating. The magnitude of the work done during this process is the integral of the force over the displacement:

$$W = \int_0^2 \mathbf{F}_m \cdot d\mathbf{x} = \int_0^2 9.0 	imes 10^{-3} e^{-0.2x} dx$$

Let's evaluate this exponential integral. The antiderivative of $e^{-0.2x}$ is $\frac{e^{-0.2x}}{-0.2}$. Applying the limits from 0 to 2:

$$W = 9.0 	imes 10^{-3} \left[ \frac{e^{-0.2x}}{-0.2} ight]_0^2$$

$$W = \frac{9.0 	imes 10^{-3}}{-0.2} (e^{-0.4} - e^0)$$

$$W = -45 	imes 10^{-3} (e^{-0.4} - 1) = 45 	imes 10^{-3} (1 - e^{-0.4}) 	ext{ J}$$

### Power: The Final Sprint

Power is the rate at which work is done, calculated as $P = \frac{W}{t}$. We are given the time $t = 5 	imes 10^{-3}$ s.

$$P = \frac{45 	imes 10^{-3} (1 - e^{-0.4})}{5 	imes 10^{-3}}$$

The $10^{-3}$ terms cancel out perfectly, leaving:

$$P = 9 (1 - e^{-0.4}) 	ext{ W}$$

To find the final numerical value, we evaluate $e^{-0.4}$. Using standard approximations or a calculator, $e^{-0.4} \approx 0.6703$.

$$P = 9 (1 - 0.6703) = 9 	imes 0.3297 \approx 2.967 	ext{ W}$$

Rounding to two decimal places, the power required is **$2.97$ W**, which perfectly matches option (b).

Select Answer:

Visualized Solution

The Physical Setup

Magnetic Force on a Current-Carrying Wire

Calculating the Magnetic Force

Direction of the Magnetic Force

Work Done by the Force

Integrating the Work Done

Calculating Average Power

Final Numerical Value

The Sigma Insight: Magnetic Force on Current

Visualizing the 3D Setup

The Lorentz Force in Action

The Calculus of Work

Power

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