Faraday's law states that an emf is induced in a coil when the magnetic flux through the coil changes with time.
Magnetic flux (\( \Phi \)) is given by:
\( \Phi = \vec{B} \cdot \vec{A} = BA \cos \theta \)
where \( \vec{B} \) is the magnetic field, \( \vec{A} \) is the area vector of the coil, and \( \theta \) is the angle between the magnetic field and the area vector.
An induced emf can be produced by rotating the coil (C) and by changing the area of the coil (D) (Option 2).
Induced emf can be induced in a coil by changing magnetic flux. And 𝜙 = 𝐵⃗ . 𝑑𝐴⃗⃗⃗⃗⃗ By rotating coil, angle between coil and magnetic field changes and hence flux changes. By changing area, magnetic flux changes.
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.}
Let A = \(\begin{bmatrix} \log_5 128 & \log_4 5 \log_5 8 & \log_4 25 \end{bmatrix}\) \). If \(A_{ij}\) is the cofactor of \( a_{ij} \), \( C_{ij} = \sum_{k=1}^2 a_{ik} A_{jk} \), and \( C = [C_{ij}] \), then \( 8|C| \) is equal to:
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.