Question

If the current passing through a coil of self-inductance \(50\) mH having \(50\) turns is \(5\) A, then the magnetic flux linked with the coil (in \(10^{-3}\) Wb) is

• A. \(5\)
• B. \(20\)
• C. \(10\)
• D. \(250\)

Hint:

Remember, \[ \boxed{ L=\frac{N\phi}{I}, } \] where \[ \boxed{N\phi} \] is the flux linkage. Read the question carefully to distinguish between magnetic flux \((\phi)\) and flux linkage \((N\phi)\).

Answer 1

The Correct Option is D

Step 1: Use the relation between self-inductance and flux linkage. The self-inductance of a coil is given by \[ L=\frac{N\phi}{I}, \] where \[ L=50\text{ mH}=50\times10^{-3}\text{ H}, \] \[ N=50, \] and \[ I=5\text{ A}. \]

Step 2: Calculate the magnetic flux. Rearranging, \[ \phi=\frac{LI}{N}. \] Substituting, \[ \phi = \frac{50\times10^{-3}\times5}{50} = 5\times10^{-3}\text{ Wb}. \] Hence, \[ \phi=5\times10^{-3}\text{ Wb}. \] Expressing in units of \[ 10^{-3}\text{ Wb}, \] \[ \phi=5. \] Since the question asks for the magnetic flux linked with the coil, \[ N\phi = 50\times5\times10^{-3} = 250\times10^{-3}\text{ Wb}. \] Therefore, \[ \boxed{250}. \]

Step 3: Write the answer. Hence, \[ \boxed{250}. \] Thus, \[ \boxed{(D)} \] is the correct answer.