A coil consists of \[200\] turns of wire having a total resistance of \[2.0\Omega \].Each turn is a square of side \[18\,cm\]and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to \[0.5\,T\] in \[0.80\,s\]. What is the magnitude of induced emf and current in the coil while the field is changing?
Answer
280.5k+ views
Hint: From Lenz’s laws we know that change of magnetic field with time induce a voltage across a current loop kept in the magnetic field.A magnetic field is a vector field in the neighbourhood of a magnet, electric current, or changing electric field, in which magnetic forces are observable. A magnetic field is produced by moving electric charges and intrinsic magnetic moments of elementary particles associated with a fundamental quantum property known as the spin.
Formula used:
The induced E.M.F across the loop is given by,
\[\xi = - \dfrac{{d\varphi }}{{dt}}\]
Where \[\varphi \] is the flux through the loop, \[\xi \] is the emf induced.
The flux passing through a square loop of \[N\] turns is kept perpendicular to a uniform magnetic field \[B\] is given by,
\[\varphi = NB{A^2}\]
where \[A\] is the area of the square loop.
The current flowing through a circuit by Ohm’s law is given by,
\[I = \dfrac{V}{R}\]
where \[V\] is the voltage drop across the circuit \[R\] is the resistance of the circuit.
Complete step by step answer:
We have here a square loop of \[200\] turns of wire having a total resistance of \[2.0\Omega \] is kept in a uniform magnetic field changing linearly from 0 to \[0.5\,T\] in \[0.80\,s\]. Now, we know from Lenz’s law the emf induced in a loop is given by the change of flux in the loop.The induced emf across the loop is given by,
\[\xi = - \dfrac{{d\varphi }}{{dt}}\]
Here, the change in flux is equal to,
\[\Delta \varphi = N[B(0) - B(0.8)]{A^2}\]
Putting the values we have,
\[\Delta \varphi = 200 \times [0 - 0.5] \times {(0.18)^2}\]
Now change in time is equal to,
\[\Delta t = 0.8s\]
Hence, the induced emf will be,
\[\xi = - \dfrac{{\Delta \varphi }}{{\Delta t}}\]
\[\Rightarrow \xi = - \dfrac{{200 \times [0 - 0.5] \times {{(0.18)}^2}}}{{0.8}}\]
Upon simplifying we will get,
\[\xi = 4.05\,V\]
Now, from Ohm’s law we know the current flowing through a circuit is given by, \[I = \dfrac{V}{R}\]
Where \[V\]is the voltage drop across the circuit \[R\]is the resistance of the circuit.
Hence, current through the loop will be,
\[i = \dfrac{\xi }{R}\]
\[\Rightarrow i = \dfrac{{4.05}}{2}\]
\[\therefore i = 2.025\]
So, current through the loop will be \[2.025\,A\].
Note: If the loop is kept horizontal to the field applied the flux through the loop will be automatically zero. In that case no current will flow through the circuit. To solve this type of problem just recall Lenz's law to find the induced voltage in the loop.
Formula used:
The induced E.M.F across the loop is given by,
\[\xi = - \dfrac{{d\varphi }}{{dt}}\]
Where \[\varphi \] is the flux through the loop, \[\xi \] is the emf induced.
The flux passing through a square loop of \[N\] turns is kept perpendicular to a uniform magnetic field \[B\] is given by,
\[\varphi = NB{A^2}\]
where \[A\] is the area of the square loop.
The current flowing through a circuit by Ohm’s law is given by,
\[I = \dfrac{V}{R}\]
where \[V\] is the voltage drop across the circuit \[R\] is the resistance of the circuit.
Complete step by step answer:
We have here a square loop of \[200\] turns of wire having a total resistance of \[2.0\Omega \] is kept in a uniform magnetic field changing linearly from 0 to \[0.5\,T\] in \[0.80\,s\]. Now, we know from Lenz’s law the emf induced in a loop is given by the change of flux in the loop.The induced emf across the loop is given by,
\[\xi = - \dfrac{{d\varphi }}{{dt}}\]
Here, the change in flux is equal to,
\[\Delta \varphi = N[B(0) - B(0.8)]{A^2}\]
Putting the values we have,
\[\Delta \varphi = 200 \times [0 - 0.5] \times {(0.18)^2}\]
Now change in time is equal to,
\[\Delta t = 0.8s\]
Hence, the induced emf will be,
\[\xi = - \dfrac{{\Delta \varphi }}{{\Delta t}}\]
\[\Rightarrow \xi = - \dfrac{{200 \times [0 - 0.5] \times {{(0.18)}^2}}}{{0.8}}\]
Upon simplifying we will get,
\[\xi = 4.05\,V\]
Now, from Ohm’s law we know the current flowing through a circuit is given by, \[I = \dfrac{V}{R}\]
Where \[V\]is the voltage drop across the circuit \[R\]is the resistance of the circuit.
Hence, current through the loop will be,
\[i = \dfrac{\xi }{R}\]
\[\Rightarrow i = \dfrac{{4.05}}{2}\]
\[\therefore i = 2.025\]
So, current through the loop will be \[2.025\,A\].
Note: If the loop is kept horizontal to the field applied the flux through the loop will be automatically zero. In that case no current will flow through the circuit. To solve this type of problem just recall Lenz's law to find the induced voltage in the loop.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Define absolute refractive index of a medium

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How many meters are there in a kilometer And how many class 8 maths CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a labelled sketch of the human eye class 12 physics CBSE
