The power dissipated in the loop can be calculated using the formula:
\[ P = I^2 R \]
Where \(I\) is the induced current and \(R\) is the resistance.
The magnetic flux change through the loop is given by:
\[ \frac{dB}{dt} = 10^{-7} \, \text{T/s} \]
Using Faraday’s Law, the induced emf in the loop is:
\[ \mathcal{E} = -N \frac{d\Phi}{dt} \]
Now, calculate the induced current \(I\):
\[ I = \frac{\mathcal{E}}{R} \]
Substituting the values, we find the power dissipated:
\[ P = 2.16 \times 10^{-9} \, \text{W} \]
\[ P = 216 \times 10^{-9} \, \text{W} \]
The problem requires us to find the power dissipated as heat in a rectangular loop. The loop is moving with a constant velocity in a region where the magnetic field varies with both position and time. This means there will be two sources of induced electromotive force (EMF).
The total induced EMF in the loop is the sum of the EMF due to the time-varying magnetic field (transformer EMF) and the motional EMF due to the loop's movement through a spatially varying magnetic field.
1. Transformer EMF (\(\varepsilon_t\)): This is induced when the magnetic flux through a stationary loop changes with time, according to Faraday's law of induction.
\[ |\varepsilon_t| = \left| -A \frac{dB}{dt} \right| = A \left| \frac{dB}{dt} \right| \]
where \(A\) is the area of the loop and \(\frac{dB}{dt}\) is the rate of change of the magnetic field with time.
2. Motional EMF (\(\varepsilon_m\)): This is induced when a conductor moves through a magnetic field. For a loop moving in a field with a spatial gradient, the EMF is given by:
\[ |\varepsilon_m| = A v \left| \frac{dB}{dx} \right| \]
where \(v\) is the velocity of the loop and \(\frac{dB}{dx}\) is the magnetic field gradient along the direction of motion.
3. Lenz's Law: This law determines the direction of the induced current. The induced current will flow in a direction that opposes the change in magnetic flux that produced it.
4. Power Dissipation: The power dissipated as heat in a loop with resistance \(R\) and total induced EMF \(\varepsilon_{total}\) is given by:
\[ P = \frac{\varepsilon_{total}^2}{R} \]
Step 1: List the given parameters and convert them to SI units.
Length of the loop, \(l = 12 \, \text{cm} = 0.12 \, \text{m}\).
Width of the loop, \(b = 5 \, \text{cm} = 0.05 \, \text{m}\).
Velocity of the loop, \(v = 5 \, \text{cm/s} = 0.05 \, \text{m/s}\) (in the +x direction).
Magnetic field gradient, \(|\frac{dB}{dx}| = 10^{-3} \, \text{T/cm} = \frac{10^{-3} \, \text{T}}{10^{-2} \, \text{m}} = 0.1 \, \text{T/m}\). Since the gradient is along the negative x-direction, \(\frac{dB}{dx} = -0.1 \, \text{T/m}\).
Rate of change of magnetic field with time, \(|\frac{dB}{dt}| = 10^{-3} \, \text{T/s}\). Since the field is decreasing, \(\frac{dB}{dt} = -10^{-3} \, \text{T/s}\).
Resistance of the loop, \(R = 6 \, \text{m}\Omega = 6 \times 10^{-3} \, \Omega\).
Area of the loop, \(A = l \times b = 0.12 \, \text{m} \times 0.05 \, \text{m} = 0.006 \, \text{m}^2 = 6 \times 10^{-3} \, \text{m}^2\).
Step 2: Determine the direction of induced EMFs using Lenz's Law.
The magnetic field is in the +z direction.
Since both effects drive the current in the same direction, their corresponding EMFs will add up.
Step 3: Calculate the magnitude of the transformer EMF (\(\varepsilon_t\)).
\[ |\varepsilon_t| = A \left| \frac{dB}{dt} \right| = (6 \times 10^{-3} \, \text{m}^2) \times (10^{-3} \, \text{T/s}) \] \[ |\varepsilon_t| = 6 \times 10^{-6} \, \text{V} \]
Step 4: Calculate the magnitude of the motional EMF (\(\varepsilon_m\)).
\[ |\varepsilon_m| = A v \left| \frac{dB}{dx} \right| = (6 \times 10^{-3} \, \text{m}^2) \times (0.05 \, \text{m/s}) \times (0.1 \, \text{T/m}) \] \[ |\varepsilon_m| = (6 \times 10^{-3}) \times (5 \times 10^{-2}) \times (10^{-1}) \, \text{V} = 30 \times 10^{-6} \, \text{V} \]
Step 5: Calculate the total induced EMF (\(\varepsilon_{total}\)).
As determined in Step 2, the EMFs add up.
\[ \varepsilon_{total} = |\varepsilon_t| + |\varepsilon_m| = 6 \times 10^{-6} \, \text{V} + 30 \times 10^{-6} \, \text{V} \] \[ \varepsilon_{total} = 36 \times 10^{-6} \, \text{V} \]
Now, we calculate the power dissipated by the loop as heat using the total EMF and the resistance.
\[ P = \frac{\varepsilon_{total}^2}{R} = \frac{(36 \times 10^{-6} \, \text{V})^2}{6 \times 10^{-3} \, \Omega} \] \[ P = \frac{1296 \times 10^{-12} \, \text{V}^2}{6 \times 10^{-3} \, \Omega} \] \[ P = 216 \times 10^{-9} \, \text{W} \]
The power dissipated by the loop as heat is 216 \( \times 10^{-9}\) W.

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.