A horizontal straight wire 5 m$5\text{ m}$ long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field 0.60 × 10⁻⁴ Wb m⁻²$0.60 \times 10^{-4}\text{ Wb m}^{-2}$. The instantaneous value of emf induced in the wire when its velocity is 10 ms⁻¹$10\text{ ms}^{-1}$ is x × 10⁻³ V$x \times 10^{-3}\text{ V}$. The value of x$x$ is:

Related Formula

Core Logic

Calculate the time derivative of the magnetic flux to find the induced EMF. Evaluate it at the requested time, and then apply Ohm's law to find the current magnitude.

Step 1: Calculate Induced EMF

At t = 2 s$t = 2 \text{ s}$:

Step 2: Calculate Induced Current

Pattern Recognition

Flux polynomials (At² - Bt + C$At^2 - Bt + C$) instantly trigger a simple derivative test. Remember to drop the negative sign for final current magnitude unless direction is specifically asked.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Related Formula

Core Logic

Assume a current i$i$ flows through the larger square loop of side L$L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side $\ell$.

The magnetic field at the center of the large square loop (distance d = L/2$d = L/2$ from each side, angles 45^°$45^\circ$):

Step 2: Mutual Inductance Calculation

Flux linkage for the inner loop:

Given L = ²$L = \ell^2$:

Mutual inductance M$M$:

Using μ₀ = 4π × 10⁻⁷$\mu_0 = 4\pi \times 10^{-7}$:

Comparing with √(x) × 10⁻⁷$\sqrt{x} \times 10^{-7}$, we get x = 128$x = 128$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Related Formula

Core Logic

Given data: Initial Magnetic Field, Bᵢ = 5000 T$B_i = 5000\mathrm{\,T}$ Final Magnetic Field, Bf = 3000 T$B_f = 3000\mathrm{\,T}$ Time interval, Δ t = 2 s$\Delta t = 2\mathrm{\,s}$ Diameter, d = 0.02 m ⇒ r = 0.01 m$d = 0.02\mathrm{\,m} \Rightarrow r = 0.01\mathrm{\,m}$ Induced emf, ε = 22 V$\varepsilon = 22\mathrm{\,V}$

Change in magnetic field magnitude |Δ B| = 5000 - 3000 = 2000 T$|\Delta B| = 5000 - 3000 = 2000\mathrm{\,T}$. Area of the coil A = π r² = π (0.01)² = 10⁻⁴π m²$A = \pi r^2 = \pi (0.01)^2 = 10^{-4}\pi \mathrm{\,m^2}$.

Step 2: Equation Evaluation

Using Faraday's Law:

22 = N (0.1π)$22 = N (0.1\pi)$ Taking π ≈ 22/7$\pi \approx 22/7$:

Chapter Mix

Class 12 Physics: Electromagnetic Induction