
A 0.5m long solenoid of 10turns/cm has an area of cross section $1c{m^2}$. Calculate the voltage induced across its ends if the current in the solenoid is charged from $1A$ to $2A$ in $0.1\sec $.
Answer
136.2k+ views
Hint: The voltage induced in the solenoid is directly proportional to the number of turns of coil and change in magnetic flux by applying Faraday's second law of electromagnetic induction.
Formula used:
Change in magnetic field, $\Delta B = {\mu _0}n\Delta I$
Where, $n = $ Number of turns per unit length; Turn density
$I = $ Current in the coil
${\mu _0} = $ Magnetic Permeability in free space
EMF or voltage induced, $\varepsilon = N\dfrac{{d\phi }}{{dt}}$
Where, $N = $ Number of turns in solenoid
$\phi = B \times A$
Complete step by step solution:
A solenoid consists of a straight coil with many turns to produce a nearly uniform magnetic field inside. The direction of the magnetic field within the coil depends upon the direction of current.
The magnetic field induced in a long solenoid is given the formula:
$B = \mu nI$
where $n$ = Number of turns per unit length; Turn density
$I$ = Current in the coil
$\mu $ = Magnetic Permeability
We are given that the Changing current, $\Delta I = 2 - 1 = 1A$;
Changing time, $\Delta t = 0.1\sec $; Area of cross section, $A = 1 \times {10^{ - 4}}{m^2}$, Number of turns, $N = 50 \times 10 = 500$
Therefore, the required magnetic field, $B = \mu n\Delta I$
$ \Rightarrow B = 4 \times 3.14 \times {10^{ - 4}} \times 10 \times 1$
Which gives, $B = 1.25 \times {10^{ - 5}}T$
Now for calculating induced EMF we can put the value of $B$ in formula 2 (in formulas used) which gives $\varepsilon = N\dfrac{{d\phi }}{{dt}}$. Where, $N = $ Number of turns in solenoid, $\phi = B \times A$
That is, $\varepsilon = N\dfrac{{B \times A}}{{dt}}$
Putting the values, $\varepsilon = 500 \times \dfrac{{1.25 \times {{10}^{ - 5}} \times 1 \times {{10}^{ - 4}}}}{{0.1}} = 500 \times 1.25 \times {10^{ - 10}}$
Therefore, $\varepsilon = 6.25 \times {10^{ - 6}}V.$
Hence the final induced emf across the ends is $6.25\mu V$. This will be our final answer.
Note: A solenoid is like an electromagnet which is used to convert electric current into mechanical forces. The magnetic field within the solenoid depends on the density of turns and current passing through it. A solenoid is the permanent magnet that can be turned off and on with will.
Formula used:
Change in magnetic field, $\Delta B = {\mu _0}n\Delta I$
Where, $n = $ Number of turns per unit length; Turn density
$I = $ Current in the coil
${\mu _0} = $ Magnetic Permeability in free space
EMF or voltage induced, $\varepsilon = N\dfrac{{d\phi }}{{dt}}$
Where, $N = $ Number of turns in solenoid
$\phi = B \times A$
Complete step by step solution:
A solenoid consists of a straight coil with many turns to produce a nearly uniform magnetic field inside. The direction of the magnetic field within the coil depends upon the direction of current.
The magnetic field induced in a long solenoid is given the formula:
$B = \mu nI$
where $n$ = Number of turns per unit length; Turn density
$I$ = Current in the coil
$\mu $ = Magnetic Permeability
We are given that the Changing current, $\Delta I = 2 - 1 = 1A$;
Changing time, $\Delta t = 0.1\sec $; Area of cross section, $A = 1 \times {10^{ - 4}}{m^2}$, Number of turns, $N = 50 \times 10 = 500$
Therefore, the required magnetic field, $B = \mu n\Delta I$
$ \Rightarrow B = 4 \times 3.14 \times {10^{ - 4}} \times 10 \times 1$
Which gives, $B = 1.25 \times {10^{ - 5}}T$
Now for calculating induced EMF we can put the value of $B$ in formula 2 (in formulas used) which gives $\varepsilon = N\dfrac{{d\phi }}{{dt}}$. Where, $N = $ Number of turns in solenoid, $\phi = B \times A$
That is, $\varepsilon = N\dfrac{{B \times A}}{{dt}}$
Putting the values, $\varepsilon = 500 \times \dfrac{{1.25 \times {{10}^{ - 5}} \times 1 \times {{10}^{ - 4}}}}{{0.1}} = 500 \times 1.25 \times {10^{ - 10}}$
Therefore, $\varepsilon = 6.25 \times {10^{ - 6}}V.$
Hence the final induced emf across the ends is $6.25\mu V$. This will be our final answer.
Note: A solenoid is like an electromagnet which is used to convert electric current into mechanical forces. The magnetic field within the solenoid depends on the density of turns and current passing through it. A solenoid is the permanent magnet that can be turned off and on with will.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

How to find Oxidation Number - Important Concepts for JEE

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Collision - Important Concepts and Tips for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

A given ray of light suffers minimum deviation in an class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Other Pages
Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

A transformer is used to light a 100W and 110V lamp class 12 physics JEE_Main

Functional Equations - Detailed Explanation with Methods for JEE

Two straight infinitely long and thin parallel wires class 12 physics JEE_Main

A point object O is placed at distance of 03 m from class 12 physics JEE_MAIN

AB is a long wire carrying a current I1 and PQRS is class 12 physics JEE_Main
