Question

The self inductance of a coil having 500 turns is 50mH. The magnetic flux through the cross-sectional area of the coil while current through it is 8mA, is found to be?
$A.\quad 4\times {{10}^{-4}}Wb$
$B.\quad 0.04Wb$
$C.\quad 4\mu Wb$
$D.\quad 40mWb$

Answer 1

Hint: The concept of self inductance in a coil is required to solve this problem. The different formulas used in the concept of inductance such as induced EMF, ${{E}_{ind}}=\dfrac{-d{{\phi }_{m}}}{dt}$ and the flux through the coil having each loops is, ${{\phi }_{m}}=\dfrac{LI}{N}$ and flux through the cross sectional area of the loop is, $\phi =LI$.

Step by step solution:
Let’s start by understanding the concept of self inductance. In mutual inductance, the change in magnetic field of a primary coil induces a secondary EMF in a secondary coil not containing any potential source. However, when the current in the primary coil changes with time, then the magnetic flux through the coil would change with time too. This would cause an EMF to be induced in the loop. This phenomenon is known as self –induction.

Let’s consider the magnetic flux produced by the coil is\[({{\phi }_{m}})\], then the amount of EMF induced$({{E}_{ind}})$ due to it, in time dt is given by, ${{E}_{ind}}=\dfrac{-d{{\phi }_{m}}}{dt}.$

The magnetic flux due to the magnetic field being produced is directly proportional to the current (I) flowing through the circuit. That is,${{\phi }_{m}}\propto I.$

This changes to, ${{\phi }_{m}}=\dfrac{LI}{N},$where (L), the self inductance is the constant of proportionality. N are the number of turns in the coil.

Hence, the final equation for the EMF induced changes to, ${{E}_{ind}}=\dfrac{-LdI}{dt}.$

In the current problem given are, \[L=50mH=50\times {{10}^{-3}}H\], number of turns in the coil, N=500 and the amount of current flowing through the coil is \[I=8mA=8\times {{10}^{-3}}A\].

Hence, the amount of magnetic flux through the cross sectional area is, $\phi =N{{\phi }_{m}}=N(\dfrac{LI}{N})=500\times \dfrac{(50\times {{10}^{-3}})(8\times {{10}^{-3}})}{500}=400\times {{10}^{-6}}Wb=4\times {{10}^{-4}}Wb.$
Hence, \[\phi =4\times {{10}^{-4}}Wb.\]

Note:
The amount of EMF induced $({{E}_{ind}})$ due to the magnetic flux through each loop \[({{\phi }_{m}})\] produced by the coil having N number of turns in it, in time dt is, ${{E}_{ind}}=\dfrac{-Nd{{\phi }_{m}}}{dt}.$ The magnetic flux \[({{\phi }_{m}})\] is also equal to, ${{\phi }_{m}}=\dfrac{LI}{N}.$ Hence, form the above equations, the final EMF induced becomes, ${{E}_{ind}}=\dfrac{-LdI}{dt}.$