Problem 89 - Master Resource Book in JEE Main Physics

Question

An electric field is given by \(\mathbf{E}=(y \hat{\mathbf{i}}+x \hat{\mathbf{j}}) \mathrm{NC}^{-1}\). The work done in moving a \(1 \mathrm{C}\) charge from \(\mathbf{r}_{A}=(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}})\) \(\mathrm{m}\) to \(\mathbf{r}_{B}=(4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{m}\) is (a) \(2 y\) (b) \(3 y\) (c) zero (d) infinity

Step-by-Step Solution

Answer

The work done, 4 J, doesn't match the given options, indicating a solution interpretation or entry issue.

Step 1: Understand the Problem

We have an electric field given by \( \mathbf{E} = (y \hat{\mathbf{i}} + x \hat{\mathbf{j}}) \, \mathrm{N/C} \). A charge of 1 C moves from the position \( \mathbf{r}_{A} = (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) \, \mathrm{m} \) to \( \mathbf{r}_{B} = (4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) \, \mathrm{m} \). We need to calculate the work done.

Step 2: Determine Path of Movement

The charge moves linearly from \( (2, 2) \) to \( (4, 2) \). This means the movement takes place along the x-axis while y remains constant at \(y = 2\).

Step 3: Find the Expression for Work Done

In a varying electric field, the work done \( W \) in moving a charge \( q \) from \( A \) to \( B \) is given by \( W = q \int_{A}^{B} \mathbf{E} \cdot d\mathbf{r} \), where \( d\mathbf{r} = dx \hat{\mathbf{i}} + dy \hat{\mathbf{j}} \).

Step 4: Simplify the Line Integral

Substitute \( d\mathbf{r} \) with only the x-component since y is constant (\( dy = 0 \)): \( W = q \int_{2}^{4} (y \hat{\mathbf{i}} + x \hat{\mathbf{j}}) \cdot dx \hat{\mathbf{i}} \). The dot product gives \( W = q \int_{2}^{4} y \, dx \).

Step 5: Evaluate the Integral

Since y is constant, \( y = 2 \), the integral simplifies to \( W = q \times y \times \int_{2}^{4} dx = q \times 2 \times (4 - 2) \).

Step 6: Calculate the Work Done

Using \( q = 1 \, \mathrm{C} \), we get \( W = 1 \times 2 \times 2 = 4 \, \mathrm{J} \).

Step 7: Compare to Multiple Choice Answers

The calculated work done, \( W = 4 \, \mathrm{J} \), does not match any of the given options directly, as the options are expressed in terms of \( y \), not in explicit values. However, reviewing calculations, under the simplification assumptions—no correction applies here—the work computation holds but was expected to solve initially symbolically under the problem formatting.

Key Concepts

Work Done in Electric Field

The work done in an electric field refers to the energy required to move a charge from one point to another within that field. In the given exercise, we're tasked with moving a 1 C charge from point \( \mathbf{r}_A \) to \( \mathbf{r}_B \) in a specified electric field. The electric field is vectorial, represented by \( \mathbf{E} = (y \hat{\mathbf{i}} + x \hat{\mathbf{j}}) \, \mathrm{N/C} \).To calculate the work done, we use the formula:\[ W = q \int_{A}^{B} \mathbf{E} \cdot d\mathbf{r} \]Here, \( q \) is the charge, and \( d\mathbf{r} \) is the differential displacement vector. For a path taken in this field, the work done is associated with how the field lines interact with the charge along its path. Calculating work using this method considers both magnitude and direction of the electric field along the path.In this specific problem, y remains constant during the movement, simplifying the integral. Without any change in direction along the y-axis (\( dy = 0 \)), only the x-component contributes significantly to the work done, making the computation easier.

Line Integral in Physics

A line integral in physics, especially in the context of fields, measures the work done by a force field in moving a point along a path. Here, the path extends from \(\mathbf{r}_A\) to \(\mathbf{r}_B\), and the electric force field is represented by \(\mathbf{E} = (y \hat{\mathbf{i}} + x \hat{\mathbf{j}}) \, \mathrm{N/C}\).The line integral is calculated using:\[ \int_{A}^{B} \mathbf{E} \cdot d\mathbf{r} \]A few key points about line integrals:

• They compute cumulative effects along a path, considering both force magnitude and direction.
• Offer a way to compute work done or energy transferred within vector fields.
• In electric fields, these integrals are crucial in understanding how charges interact as they traverse paths.

In this problem, the integral simplifies because the movement is restricted to the x-component only, making the computations straightforward. The integral evaluates over x from 2 to 4, taking into account constant y, which is typical in piecewise line integrals.

Motion in Electric Field

Motion in an electric field describes how a charged particle moves under the influence of the field's force. In this exercise, the charge moves in a specified electric field characterized by components \( y \hat{\mathbf{i}} \) and \( x \hat{\mathbf{j}} \).The trajectory defined from \( (2, 2) \) to \( (4, 2) \) indicates:

• The motion is linear along the x-axis.
• The y-component of motion remains zero, meaning there is no displacement in the y-direction.
• This simplifies the work calculation as only one directional component needs consideration.

Understanding motion in electric fields involves appreciating how electric forces affect charge trajectories. It's notable that in stationary fields where only one directional movement occurs, as in this example, we can simplify work and energy considerations to linear treatments.