
The magnetic field in a coil of 100 turns and 40 square cm area is increased from 1 Tesla to 6 Tesla in 2 seconds. The magnetic field is perpendicular to the coil. The e.m.f. generated in it is:
A) $104 V$
B) $1.2 V$
C) $1.0 V$
D) $10^{-2}V$
Answer
144.9k+ views
Hint: Faraday’s discovery in 1831 of the phenomenon of magnetic induction is one of the great milestones in the quest toward understanding and exploiting nature. Faraday found that a changing magnetic field in a circuit induces an electro-motive force in the circuit. Secondly, the magnitude of the electromotive force equals the rate at which the flux of the magnetic field through the circuit changes. The flux is a measure of how much field penetrates through the circuit. The electromotive force is measured in volts and is represented by the equation:
\[e = - \dfrac{{d\phi }}{{dt}}\]
Here, $\phi$ the flux of the vector field B through the circuit, measures how much of the field passes through the circuit. The rate of change of this flux is the induced electromotive force. The units of magnetic flux are Weber. The minus sign indicates the direction of the induced electromotive force and hence of any induced current.
Complete step by step solution:
The number of turns of the coil (N) = 100.
Area of the cross-section of the coil (A) = 40 cm2.
Initial magnetic flux of the coil (\[{\phi _1}\]) = 1 Tesla.
Final magnetic flux of the coil (\[{\phi _2}\]) = 6 Tesla.
Change in the magnetic flux of the coil (\[{\phi _2} - {\phi _1}\]) = 6 – 1= 5 Tesla.
The time required to change in the magnetic flux (t) = 2 s.
Now by applying Faraday Second Law of the Electromagnetic Induction, we get
$
\Rightarrow e = - N\dfrac{{d\phi }}{{dt}} \\
\Rightarrow \phi = \overrightarrow B .\overrightarrow A = BA{{ }}\cos \theta$
Since the magnetic field $(\overrightarrow B )$ and $(\overrightarrow A )$ are parallel to each other, the angle between them is zero.
$ \\
\Rightarrow \phi = \overrightarrow B .\overrightarrow A = BA{{ }}\cos 0^\circ \\
\Rightarrow e = - N\dfrac{{dBA{{ }}\cos 0^\circ }}{{dt}} \\
\Rightarrow e = - NA{{ }}\cos 0^\circ \dfrac{{dB}}{{dt}} \\
\Rightarrow e = - NA{{ }}\cos 0^\circ \dfrac{{d({\phi _2} - {\phi _1})}}{{dt}} \\
\Rightarrow e = - 100 \times {{ }}\cos 0^\circ \times \dfrac{{40 \times {{10}^{ - 4}}}}{2} \\
\Rightarrow e = - 100 \times {{ }}1 \times \dfrac{{40 \times {{10}^{ - 4}}}}{2} \\
\Rightarrow e = - 100 \times {{ }}1 \times 20 \times {10^{ - 4}} \times (6 - 1) \\
\Rightarrow e = - 100 \times {{ }}1 \times 20 \times {10^{ - 4}} \times 5 \\
\Rightarrow e = - 100 \times {{ }}1 \times 100 \times {10^{ - 4}} \\
\Rightarrow e = - 1{{ }}V \\
\Rightarrow \left| e \right| = 1{{ }}V. \\
$
The electromotive force generated in the coil is 1 V.
Therefore, option(C) is correct.
Note: Faraday found that a changing magnetic field in a circuit induces an electro-motive force in the circuit. Secondly, the magnitude of the electromotive force equals the rate at which the flux of the magnetic field through the circuit changes. The flux is a measure of how much field penetrates through the circuit. The electromotive force is measured in volts and is represented by the equation:
$
e = - N\dfrac{{d\phi }}{{dt}} \\
\Rightarrow e = - NA\dfrac{{dB}}{{dt}}\cos \theta . \\
$
\[e = - \dfrac{{d\phi }}{{dt}}\]
Here, $\phi$ the flux of the vector field B through the circuit, measures how much of the field passes through the circuit. The rate of change of this flux is the induced electromotive force. The units of magnetic flux are Weber. The minus sign indicates the direction of the induced electromotive force and hence of any induced current.
Complete step by step solution:
The number of turns of the coil (N) = 100.
Area of the cross-section of the coil (A) = 40 cm2.
Initial magnetic flux of the coil (\[{\phi _1}\]) = 1 Tesla.
Final magnetic flux of the coil (\[{\phi _2}\]) = 6 Tesla.
Change in the magnetic flux of the coil (\[{\phi _2} - {\phi _1}\]) = 6 – 1= 5 Tesla.
The time required to change in the magnetic flux (t) = 2 s.
Now by applying Faraday Second Law of the Electromagnetic Induction, we get
$
\Rightarrow e = - N\dfrac{{d\phi }}{{dt}} \\
\Rightarrow \phi = \overrightarrow B .\overrightarrow A = BA{{ }}\cos \theta$
Since the magnetic field $(\overrightarrow B )$ and $(\overrightarrow A )$ are parallel to each other, the angle between them is zero.
$ \\
\Rightarrow \phi = \overrightarrow B .\overrightarrow A = BA{{ }}\cos 0^\circ \\
\Rightarrow e = - N\dfrac{{dBA{{ }}\cos 0^\circ }}{{dt}} \\
\Rightarrow e = - NA{{ }}\cos 0^\circ \dfrac{{dB}}{{dt}} \\
\Rightarrow e = - NA{{ }}\cos 0^\circ \dfrac{{d({\phi _2} - {\phi _1})}}{{dt}} \\
\Rightarrow e = - 100 \times {{ }}\cos 0^\circ \times \dfrac{{40 \times {{10}^{ - 4}}}}{2} \\
\Rightarrow e = - 100 \times {{ }}1 \times \dfrac{{40 \times {{10}^{ - 4}}}}{2} \\
\Rightarrow e = - 100 \times {{ }}1 \times 20 \times {10^{ - 4}} \times (6 - 1) \\
\Rightarrow e = - 100 \times {{ }}1 \times 20 \times {10^{ - 4}} \times 5 \\
\Rightarrow e = - 100 \times {{ }}1 \times 100 \times {10^{ - 4}} \\
\Rightarrow e = - 1{{ }}V \\
\Rightarrow \left| e \right| = 1{{ }}V. \\
$
The electromotive force generated in the coil is 1 V.
Therefore, option(C) is correct.
Note: Faraday found that a changing magnetic field in a circuit induces an electro-motive force in the circuit. Secondly, the magnitude of the electromotive force equals the rate at which the flux of the magnetic field through the circuit changes. The flux is a measure of how much field penetrates through the circuit. The electromotive force is measured in volts and is represented by the equation:
$
e = - N\dfrac{{d\phi }}{{dt}} \\
\Rightarrow e = - NA\dfrac{{dB}}{{dt}}\cos \theta . \\
$
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

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

Physics Average Value and RMS Value JEE Main 2025

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