Question

If \(\vec V=(x+3y)\hat i+(y-2x)\hat j+(x+az)\hat k\) is a solenoidal, then \(a=\)

• A. \(0\)
• B. \(1\)
• C. \(-2\)
• D. \(\dfrac{1}{2}\)

Hint:

A vector field is solenoidal if its divergence is zero.

Answer 1

The Correct Option is C

Concept:
A vector field \(\vec V\) is called solenoidal if its divergence is zero. That is, \[ \nabla\cdot \vec V=0 \]

Step 1: Write the vector field. \[ \vec V=(x+3y)\hat i+(y-2x)\hat j+(x+az)\hat k \] Let \[ P=x+3y,\quad Q=y-2x,\quad R=x+az \]

Step 2: Find divergence. \[ \nabla\cdot \vec V= \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \] Now, \[ \frac{\partial}{\partial x}(x+3y)=1 \] \[ \frac{\partial}{\partial y}(y-2x)=1 \] \[ \frac{\partial}{\partial z}(x+az)=a \] Therefore, \[ \nabla\cdot \vec V=1+1+a \] \[ \nabla\cdot \vec V=2+a \]

Step 3: Use solenoidal condition. Since the vector field is solenoidal, \[ \nabla\cdot \vec V=0 \] So, \[ 2+a=0 \] \[ a=-2 \]

Step 4: Final answer. \[ \boxed{-2} \]