A rectangular loop of sides 12 cm and 5 cm, with its sides parallel to the x-axis and y-axis respectively, moves with a velocity of 5 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 along the negative x direction, and it is decreasing with time at the rate of \(10^{-3}\) T/s. If the resistance of the loop is 6 mΩ, the power dissipated by the loop as heat is \(\_\_\_\_\_\_ \times 10^{-9}\) W.
The field has a gradient of \(10^{-3}\) T/cm along the negative x direction, and it is decreasing with time at the rate of \(10^{-3}\) T/s. If the resistance of the loop is 6 mΩ, the power dissipated by the loop as heat is \(\_\_\_\_\_\_ \times 10^{-9}\) W.
Correct Answer: 216
Approach Solution - 1
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:
\[ P = 2.16 \times 10^{-9} \, \text{W} \]
\[ P = 216 \times 10^{-9} \, \text{W} \]
Approach Solution -2
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).
Concept Used:
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-by-Step Solution:
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.
- Due to time variation: The field is decreasing (\(\frac{dB}{dt} < 0\)). The flux in the +z direction is decreasing. To oppose this, the induced current will create a magnetic field in the +z direction. By the right-hand rule, this corresponds to a counter-clockwise (CCW) current.
- Due to motion: The loop moves in the +x direction, and the field has a gradient along the -x direction (\(\frac{dB}{dx} < 0\)), meaning the field strength decreases as x increases. As the loop moves into a region of weaker field, the flux in the +z direction decreases. To oppose this change, the induced current will again create a magnetic field in the +z direction, resulting in a CCW current.
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.
\[ \varepsilon_{total} = |\varepsilon_t| + |\varepsilon_m| = 6 \times 10^{-6} \, \text{V} + 30 \times 10^{-6} \, \text{V} \] \[ \varepsilon_{total} = 36 \times 10^{-6} \, \text{V} \]
Final Computation & Result:
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.
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