Question

A coil having 100 turns is carrying of 2 A. If magnetic flux linked with each turn of coil is $5\times {10}^3$ Wb then the self inductance of the coil is:

• A. ${10}^{-3}$ H
• B. $2.5\times {10}^{5}H$
• C. $4\times {10}^{-3}H$
• D. $4\times {10}^{-4}$H

Answer 1

The correct option is B $2.5\times {10}^{5}H$

Total flux linked ${\varphi }_{total}=N\times {\varphi }_{single}=100\times 5\times {10}^3=5\times {10}^{5}Wb$

Current $i=2A$

Self inductance $L=\frac{{\varphi }_{total}}{i}=\frac{5\times {10}^5}{2}=2.5\times {10}^{5}H$