A rectangular loop of sides 12~cm$12\mathrm{~cm}$ and 5~cm$5\mathrm{~cm}$, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of 5~cm/s$5\mathrm{~cm/s}$ in the positive x-axis direction, in a space containing a variable magnetic field in the positive z-direction. The field has a gradient of 10⁻³~T/cm$10^{-3}\mathrm{~T/cm}$ along the negative x-direction and it is decreasing with time at the rate of 10⁻³~T/s$10^{-3}\mathrm{~T/s}$. If the resistance of the loop is 6~mΩ$6\mathrm{~m\Omega}$, the power dissipated by the loop as heat is x × 10⁻⁹~W$x \times 10^{-9}\mathrm{~W}$. The value of x$x$ is:

Related Formula

Core Logic

The magnetic field is constant, but the area changes as the circle deforms into a square. Length of the wire is conserved. Circumference of circular loop = 2π r ≈ 14π cm$2\pi r \approx 14\pi \, \mathrm{cm}$.

Side length of the new square loop:

Step 1: Calculate Change in Area

Step 2: Calculate Flux Change and EMF

Using π ≈ 3.1415$\pi \approx 3.1415$: 49π ≈ 153.938$49\pi \approx 153.938$ (49π²)/(4) ≈ 120.89$\frac{49\pi^2}{4} \approx 120.89$ Δ A ≈ 33.04 × 10⁻⁴ m²$\Delta A \approx 33.04 \times 10^{-4} \, \mathrm{m}^2$

Wait, taking standard approximations:

Pattern Recognition

When a loop is deformed into a different shape, perimeter is conserved. For a given perimeter, a circle always encloses maximum area. Subtracting the new area from the old area immediately gives the Δ A$\Delta A$ driving the induced EMF.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Related Formula

B = μ₀ n I$B = \mu_0 n I$

Core Logic

Since the loop B is well inside the long solenoid (10 cm deep into a 100 cm length), the magnetic field is uniform across its cross-section and approximately equal to the ideal solenoid field B = μ₀ n I$B = \mu_0 n I$. Here n = 500$n = 500$ turns/m. The sliding velocity v$v$ just means it's moving through a uniform field region, so motional EMF does not contribute to the flux change; only the transformer EMF (time-varying B$B$ field) creates the induced current.

Step 1: Calculate Induced EMF

Step 2: Calculate RMS Current

Peak EMF ε₀ = π r² μ₀ n I₀ ω$\epsilon_0 = \pi r^2 \mu_0 n I_0 \omega$.

Step 3: Numerical Substitution

Using π² ≈ 9.87$\pi^2 \approx 9.87$, 20π² ≈ 197.39$20\pi^2 \approx 197.39$. Thus, irms ≈ 197√(2)$i_{\text{rms}} \approx \frac{197}{\sqrt{2}} \, \mu\mathrm{A}$. α = 197$\alpha = 197$.

Pattern Recognition

Motion parallel to uniform field lines inside an ideal solenoid induces no motional EMF. Only the time derivative of the AC current (dI)/(dt)$\frac{dI}{dt}$ contributes to Faraday induction.

Chapter Mix

Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

Related Formula

• Magnetic Energy Density (energy per unit volume):

• Total Energy stored:

Core Logic

We are given:

• Relative permeability μᵣ = 2$\mu_r = 2$
• Permeability of the filled core medium μ = μᵣ μ₀ = 2 μ₀$\mu = \mu_r \mu_0 = 2 \mu_0$

Substitute μᵣ = 2$\mu_r = 2$ into the energy density expression:

Multiply by the total volume of the solenoid (V = A l$V = A l$):

Pattern Recognition

Sees: Magnetic energy stored in a solenoid with medium relative permeability. Trap: Placing the relative permeability μᵣ$\mu_r$ in the numerator of the formula instead of the denominator. Shortcut: Magnetic energy density is always inversely proportional to permeability. With medium μ = 2μ₀$\mu = 2\mu_0$, energy density is halved compared to free space, giving (B²)/(4μ₀)$\frac{B^2}{4\mu_0}$. Multiply by volume Al$Al$ to get (B² Al)/(4μ₀)$\frac{B^2 Al}{4\mu_0}$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction Class 12 Physics: Magnetism and Matter

Related Formula

Motional electromotive force formula:

where leff$l_{\text{eff}}$ is the perpendicular component of the straight-line displacement vector connecting the endpoints A$A$ and E$E$ (lAE$l_{AE}$) relative to velocity v$\vec{v}$.

Core Logic

In a uniform magnetic field, the motional EMF induced in any arbitrary conductor wire depends solely on the straight-line displacement vector connecting its endpoints, rather than the detailed path:

Step 1: Compute Effective Length

From the geometry of the symmetric wire segments oriented at 45°$45^{\circ}$ to the horizontal:

Step 2: Calculate Induced EMF

Substitute the given parameters into the motional EMF expression:

Pattern Recognition

In a uniform magnetic field, motional EMF is path-independent. Replace any zig-zag or curved conductor with an equivalent straight line joining the two endpoints perpendicular to the velocity vector.

Evaluation Rubric / Model Answer

10

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Related Formula

The induced electromotive force (EMF) developed between the center and the rim of a rotating disc in a perpendicular magnetic field is given by:

Core Logic

Given parameters from the problem statement [cite: 655, 657, 658]:

• Magnetic field, B = 0.4 T$B = 0.4 \text{ T}$
• Radius of the disc, R = 20 cm = 0.2 m$R = 20 \text{ cm} = 0.2 \text{ m}$
• Angular velocity, \omega = 10\pi \text{ rad s}^{-1}

Substituting the values into the governing formula:

Step 1: Evaluation

The potential difference developed between the axis of the disc and the rim is precisely 0.2512 V$0.2512 \text{ V}$.

Pattern Recognition

For any rotating conductor of length R$R$ or a continuous disc rotating about its center in a perpendicular magnetic field, the induced EMF is mathematically equivalent to a single radial rod sweeping the area, leading directly to the formula (1)/(2)Bω R²$\frac{1}{2}B\omega R^2$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction