Solveeit Logo
Login

Question

Question: The magnetic field in a coil of 100 turns and 40 square cm area is increased from 1 Tesla to 6 Tesla...

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) 104V104 V
B) 1.2V1.2 V
C) 1.0V1.0 V
D) 10−2V10^{-2}V

Explanation

Solution

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=−dϕdte = - \dfrac{{d\phi }}{{dt}}
Here, ϕ\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 (ϕ1{\phi _1}) = 1 Tesla.
Final magnetic flux of the coil (ϕ2{\phi _2}) = 6 Tesla.
Change in the magnetic flux of the coil (ϕ2−ϕ1{\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
⇒e=−Ndϕdt ⇒ϕ=B→.A→=BAcos⁡θ \Rightarrow e = - N\dfrac{{d\phi }}{{dt}} \\\ \Rightarrow \phi = \overrightarrow B .\overrightarrow A = BA{{ }}\cos \theta
Since the magnetic field (B→)(\overrightarrow B ) and (A→)(\overrightarrow A ) are parallel to each other, the angle between them is zero.
 ⇒ϕ=B→.A→=BAcos⁡0∘ ⇒e=−NdBAcos⁡0∘dt ⇒e=−NAcos⁡0∘dBdt ⇒e=−NAcos⁡0∘d(ϕ2−ϕ1)dt ⇒e=−100×cos⁡0∘×40×10−42 ⇒e=−100×1×40×10−42 ⇒e=−100×1×20×10−4×(6−1) ⇒e=−100×1×20×10−4×5 ⇒e=−100×1×100×10−4 ⇒e=−1V ⇒∣e∣=1V. \\\ \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=−Ndϕdt ⇒e=−NAdBdtcos⁡θ.  e = - N\dfrac{{d\phi }}{{dt}} \\\ \Rightarrow e = - NA\dfrac{{dB}}{{dt}}\cos \theta . \\\