GATE 2009 Electromagnetic Fields Questions and Solutions

The vector field is given by:

Its line integral over the straight line from \((x, y)=(0,2)\) to \((2,0)\) evaluates to:

1. -8
2. 4
3. 8
4. 0

Solution

Detailed Steps:

1. Parametrize the line from \((0, 2)\) to \((2, 0)\): The equation of the line passing through \((0,2)\) and \((2,0)\) is \(y = -x + 2\). Let \(x = 2t\) where \(t \in [0, 1]\). Then \(y = -2t + 2 = 2(1 - t)\).
2. Calculate the derivatives:
   Equation

   \[dx = 2 dt \quad \text{and} \quad dy = -2 dt\]
3. Substitute into the vector field components:
   Equation

   \[F_x = x^2 + xy = (2t)^2 + (2t)(2(1-t)) = 4t^2 + 4t - 4t^2 = 4t\]
   Equation

   \[F_y = y^2 + xy = [2(1-t)]^2 + (2t)[2(1-t)]\]
   Equation

   \[F_y = 4(1 - 2t + t^2) + (4t - 4t^2) = 4 - 8t + 4t^2 + 4t - 4t^2 = 4 - 4t\]
4. Calculate the dot product inside the integral: The line integral is \(\int (F_x dx + F_y dy)\).
   Equation

   \[\begin{aligned} F_x dx + F_y dy &= (4t)(2 dt) + (4 - 4t)(-2 dt) \\ &= 8t \, dt + (-8 + 8t) \, dt \\ &= (16t - 8) \, dt \end{aligned}\]
5. Integrate over \(t\) from 0 to 1:
   Equation

   \[\begin{aligned} \int_{0}^{1} (16t - 8) dt &= \left[ \frac{16t^2}{2} - 8t \right]_{0}^{1} \\ &= \left[ 8t^2 - 8t \right]_{0}^{1} \\ &= (8(1)^2 - 8(1)) - (0 - 0) \\ &= 8 - 8 \\ &= 0 \end{aligned}\]