Let's go through each part step-by-step: --- ### **Part i: Sketching the Graph of Induced Current** To sketch the graph, we need some context about how the magnetic flux through the loop changes with time (which is not provided in the image, but typically for such problems, the flux changes in segments, e.g., increasing, constant, decreasing, etc.). #### **General Rules for Induced Current:** - **Faraday's Law:** The induced emf (and hence the current) in a loop is proportional to the rate of change of magnetic flux through the loop: \[ \mathcal{E} = -\frac{d\Phi_B}{dt} \] - **Lenz's Law:** The direction of the induced current opposes the change in flux. #### **Assume a Typical Pattern:** Suppose the magnetic flux through the loop: - **Increases from 0 to 8 s** - **Remains constant from 8 to 12 s** - **Decreases from 12 to 16 s** - **Remains constant from 16 to 18 s** Then: - **0–8 s:** Flux is increasing ⇒ induced current is in the direction opposing increase (say, CCW is positive, so I > 0). - **8–12 s:** Flux is constant ⇒ no change ⇒ I = 0. - **12–16 s:** Flux is decreasing ⇒ induced current opposes decrease (opposite direction to before, so I < 0, i.e., CW). - **16–18 s:** Flux constant ⇒ I = 0. #### **Graph Sketch:** - From 0 to 8 s: I is positive (CCW). - From 8 to 12 s: I = 0. - From 12 to 16 s: I is negative (CW). - From 16 to 18 s: I = 0. **You would draw a horizontal line above the time axis from 0-8 s, a line at I=0 from 8-12 s, a line below the axis from 12-16 s, and I=0 from 16-18 s.** --- ### **Part ii: Justification for 12–16 s (Current Direction)** **From 12 to 16 s:** - Magnetic flux is decreasing. - By Lenz's law, the induced current will flow in a direction that tries to maintain the original flux, i.e., it will generate a field in the same direction as the original. - If CCW was positive when flux was increasing, now the current must be **CW (negative)** to oppose the decrease. --- ### **Part c: Calculate the Total Energy Dissipated in the Loop During First 8 s** #### **Given:** We need one more piece of information: either the resistance of the loop or the value of the induced emf/current. Let's assume the current during 0–8 s is \( I_0 \) and the resistance of the loop is \( R \). #### **Energy dissipated in a resistor:** \[ E = I^2 R t \] Suppose the induced current is constant and equal to \( I_0 \) during 0–8 s. So, \[ E = (I_0)^2 R \times 8 \] **If actual values for \( I_0 \) and \( R \) are given, plug them in to get the numeric answer.** --- ## **Summary of Answers:** ### **i. Graph:** - 0–8 s: Induced current is **positive** (CCW). - 8–12 s: Induced current is **zero**. - 12–16 s: Induced current is **negative** (CW). - 16–18 s: Induced current is **zero**. ### **ii. Justification:** During 12–16 s, the magnetic flux is decreasing, so the induced current flows in the direction (CW) that tries to maintain the original flux, **opposing the decrease** (Lenz's Law). ### **c. Energy Calculation:** \[ E = (I_0)^2 R \times 8 \] Where \( I_0 \) is the current during 0–8 s and \( R \) is the resistance of the loop. --- **If you have the actual values for current and resistance, I can plug them in for you!**

Certainly! To provide detailed calculations and explanations, I need to clarify some assumptions because the image does not specify exact numerical data such as magnetic flux values, resistance, or current values. However, based on typical physics problems involving a changing magnetic flux and induced currents, I will guide you step-by-step through the process, assuming common scenarios. --- ### **Part i: Sketching the graph of induced current \( I(t) \)** #### **Understanding the problem:** - The loop experiences a change in magnetic flux over time. - The induced emf \( \mathcal{E} \) (and thus current \( I \)) depends on \( d\Phi_B/dt \). - The problem mentions a time interval from \( t = 0 \) to \( t = 18\,s \). - The axes suggest the current \( I \) is plotted with positive (CCW) and negative (CW) directions. --- ### **Step 1: Analyze the flux change (hypothetical scenario)** Suppose the magnetic flux \( \Phi_B(t) \) varies as follows: - **From 0 to 8 s:** Flux increases linearly from 0 to a maximum \( \Phi_{max} \). - **From 8 to 12 s:** Flux remains constant (no change). - **From 12 to 16 s:** Flux decreases linearly back to 0. - **From 16 to 18 s:** Flux remains constant at 0. This is a typical pattern in such problems. --- ### **Step 2: Determine the induced emf and current** - **Faraday's Law:** \[ \mathcal{E} = - \frac{d\Phi_B}{dt} \] - The induced current \( I(t) \) is related to emf via Ohm's law: \[ I(t) = \frac{\mathcal{E}}{R} \] - The sign of \( I \) depends on the direction of \( \mathcal{E} \), which is dictated by the sign of \( d\Phi_B/dt \): | Time interval | \( d\Phi_B/dt \) | \( \mathcal{E} \) sign | \( I(t) \) direction | Explanation | |-----------------|------------------|-----------------------|----------------------|--------------| | 0–8 s | Positive (flux increasing) | Negative (since \( \mathcal{E} = -d\Phi_B/dt \)) | **CCW** (positive) | Induced current opposes flux increase, but since flux increasing, the emf is negative, so the current flows CCW assuming positive is CCW. | | 8–12 s | Zero | Zero | Zero | No change in flux, no emf. | | 12–16 s | Negative (flux decreasing) | Positive (since flux decreasing, \( d\Phi_B/dt < 0 \), so emf is positive) | **CW** (negative) | Induced current flows CW to oppose flux decrease. | | 16–18 s | Zero | Zero | Zero | No flux change. | **Note:** The sign conventions depend on the initial assumptions, but following the typical right-hand rule, this pattern holds. --- ### **Step 3: Sketch the graph** - From **0 to 8 s:** \( I(t) \) is positive (say, at a constant value \( I_{max} \)). - From **8 to 12 s:** \( I(t) = 0 \). - From **12 to 16 s:** \( I(t) \) is negative (say, at \( -I_{max} \)). - From **16 to 18 s:** \( I(t) = 0 \). The graph is a rectangular pulse going up, then down, with zeros in between. --- ### **Part ii: Justify the current direction between 12 and 16 seconds** - **Flux decreasing:** The flux is decreasing during 12–16 s. - **Lenz's Law:** The induced current opposes the change — it tries to maintain the original flux. - **Result:** Since flux is decreasing, the induced current must generate a magnetic field **in the same direction as the original flux**. - **Direction:** To produce a magnetic field in the original direction, the current must flow **CW** (assuming CCW was positive during the flux increase). **Therefore, during 12–16 s, the induced current flows CW, which is negative in our sign convention.** --- ### **Part c: Calculate the total energy dissipated in the loop during the first 8 seconds** #### **Given Data (assumed):** - Resistance of the loop: \( R \) - Magnetic flux change: \( \Delta \Phi_B \) - Induced emf during flux change: \[ \mathcal{E} = - \frac{d\Phi_B}{dt} \] - Induced current: \[ I(t) = \frac{\mathcal{E}}{R} \] --- ### **Step 4: Find the emf and current during 0–8 s** Suppose: - \( \Phi_B(t) \) increases linearly from 0 to \( \Phi_{max} \) over 8 seconds: \[ \frac{d\Phi_B}{dt} = \frac{\Phi_{max} - 0}{8\,s} = \frac{\Phi_{max}}{8} \] - The emf: \[ \mathcal{E} = - \frac{d\Phi_B}{dt} = - \frac{\Phi_{max}}{8} \] - The induced current: \[ I(t) = \frac{\mathcal{E}}{R} = - \frac{\Phi_{max}}{8 R} \] Since the emf is negative when flux is increasing, the current flows CCW (positive in our convention). --- ### **Step 5: Calculate energy dissipated** The energy dissipated in the resistor over time \( t \) is: \[ E = \int_0^{8} I^2(t) R\, dt \] - Because \( I(t) \) is constant during this interval (assuming linear flux change and constant emf): \[ I = \frac{\mathcal{E}}{R} = \frac{\Phi_{max}}{8 R} \] - Power dissipated at each moment: \[ P = I^2 R = \left( \frac{\Phi_{max}}{8 R} \right)^2 R = \frac{\Phi_{max}^2}{64 R} \] - Total energy over 8 seconds: \[ E = P \times t = \frac{\Phi_{max}^2}{64 R} \times 8 = \frac{\Phi_{max}^2}{8 R} \] --- ### **### **Final answer:** \[ \boxed{ \text{Total energy dissipated during first 8 seconds} = \frac{\Phi_{max}^2}{8 R} } \] --- ### **Summary:** - **Graph of \( I(t) \):** Positive (CCW) from 0–8 s, zero from 8–12 s, negative (CW) from 12–16 s, zero from 16–18 s. - **Justification:** During 12–16 s, flux decreases; by Lenz's law, current flows CW to oppose decreasing flux. - **Energy dissipated:** \(\displaystyle \frac{\Phi_{max}^2}{8 R}\). --- **Note:** To get a numerical value, you need specific values for \( \Phi_{max} \) and \( R \). If you provide these, I can perform the actual calculation.

Certainly! Here's a detailed, step-by-step solution based on the typical assumptions and the data provided, structured clearly with explanations and calculations. --- ### **Part A: Sketch and Analyze the Induced Current \( I(t) \)** --- #### **1. Understanding the problem:** - The problem involves a changing magnetic flux through a loop, inducing a current \( I(t) \). - Assuming the magnetic flux varies over time in a specific pattern, leading to an induced current that can be positive (CCW) or negative (CW). - The axes on the graph are time \( t \) (from 0 to 18 seconds) and current \( I \), with positive indicating CCW current. --- #### **2. Assumed flux variation pattern:** Based on typical problems, assume the following pattern for magnetic flux \( \Phi_B(t) \): | Time Interval | Flux \( \Phi_B(t) \) | Change in flux | \( d\Phi_B/dt \) | Induced emf \( \mathcal{E} \) | Induced current \( I(t) \) | Direction of \( I(t) \) | |-----------------|----------------------|----------------|-----------------|------------------------------|----------------------------|-------------------------| | 0 – 8 s | Increasing from 0 to \( \Phi_{max} \) | + \( \Phi_{max} \) | \( \frac{\Phi_{max}}{8} \) | \(- \frac{\Phi_{max}}{8}\) | \( I = \frac{\mathcal{E}}{R} = - \frac{\Phi_{max}}{8 R} \) | **CCW** (positive) | | 8 – 12 s | Constant flux at \( \Phi_{max} \) | 0 | 0 | 0 | 0 | Zero | | 12 – 16 s | Decreasing flux from \( \Phi_{max} \) to 0 | \(- \Phi_{max} \) | \(- \frac{\Phi_{max}}{4} \) | \( + \frac{\Phi_{max}}{4} \) | \( \frac{\Phi_{max}}{4 R} \) | **CW** (negative) | | 16 – 18 s | Constant flux at 0 | 0 | 0 | 0 | 0 | Zero | **Note:** The flux change during 12–16s is assumed to be linear, decreasing from \( \Phi_{max} \) to 0 over 4 seconds, hence \( d\Phi_B/dt = - \frac{\Phi_{max}}{4} \). --- #### **3. Sketching \( I(t) \):** - **From 0 to 8 s:** Induced current is **positive** (CCW), at a constant magnitude: \[ I = - \frac{\Phi_{max}}{8 R} \] (Since emf is negative, current flows CCW). - **From 8 to 12 s:** No change in flux, so \[ I(t) = 0 \] - **From 12 to 16 s:** Flux decreases, emf positive, current flows **CW** (negative in our sign convention): \[ I = + \frac{\Phi_{max}}{4 R} \] - **From 16 to 18 s:** No change, \( I(t) = 0 \). **Graphically:** - Draw a horizontal line at \( I = - \frac{\Phi_{max}}{8 R} \) from 0 to 8 s. - Drop to zero from 8 to 12 s. - Jump to \( + \frac{\Phi_{max}}{4 R} \) from 12 to 16 s. - Return to zero from 16 to 18 s. --- ### **4. Justification of the current direction between 12–16 seconds** - During 12–16 s, flux is decreasing linearly. - According to **Lenz's Law**, the induced current opposes the change in flux. - Since flux decreases, the induced current **flows in a direction that creates a magnetic field in the original direction** (to oppose the decrease). - **If the flux was initially increasing CCW, then during decrease, the induced current flows CW**. - **Therefore, during 12–16 s, the current is CW (negative in the chosen sign convention).** --- ### **5. Calculation of the total energy dissipated in the loop during the first 8 seconds** --- #### **Assumptions:** - Magnetic flux increases linearly from 0 to \( \Phi_{max} \) over 8 seconds. - Resistance of the loop: \( R \) (value not specified; will keep symbolic). - The emf induced during flux change: \[ \mathcal{E} = - \frac{d\Phi_B}{dt} = - \frac{\Phi_{max}}{8} \] - The current during this interval: \[ I = \frac{\mathcal{E}}{R} = - \frac{\Phi_{max}}{8 R} \] - The power dissipated in the resistor at any instant: \[ P = I^2 R = \left( \frac{\Phi_{max}}{8 R} \right)^2 R = \frac{\Phi_{max}^2}{64 R} \] - Since the emf (and thus current) is constant over 8 seconds, total energy dissipated: \[ E = P \times t = \frac{\Phi_{max}^2}{64 R} \times 8 = \frac{\Phi_{max}^2}{8 R} \] --- ### **### **Final answers:** - **Graph of \( I(t) \):** - **0–8 s:** \( I(t) = - \frac{\Phi_{max}}{8 R} \) (positive CCW) - **8–12 s:** \( I(t) = 0 \) - **12–16 s:** \( I(t) = + \frac{\Phi_{max}}{4 R} \) (negative CW) - **16–18 s:** \( I(t) = 0 \) - **Direction justification (12–16 s):** The flux decreases; by Lenz's law, the induced current flows CW to oppose the decrease. - **Total energy dissipated during first 8 seconds:** \[ \boxed{ \text{Energy} = \frac{\Phi_{max}^2}{8 R} } \] --- ### **Summary:** This detailed step-by-step approach aligns with the typical physical principles and assumptions in such problems. If you have specific values for \( \Phi_{max} \) and \( R \), substitute them to compute a numerical answer.

Certainly! Here's a comprehensive, step-by-step explanation and solution without referencing a table, focusing solely on the reasoning and calculations: --- ### **Part A: Induced Current \( I(t) \) as a Function of Time** **Step 1: Understand the nature of magnetic flux change** - The magnetic flux through the loop varies with time. - Typically, in such problems, flux increases linearly over a certain interval, remains constant, then decreases, etc. - The direction of the induced current depends on how the flux changes: - If flux **increases**, the induced current flows **opposite** to the increasing flux to oppose the change (Lenz's Law). - If flux **decreases**, the induced current flows **to support** the original flux, i.e., in the same direction as before. **Step 2: Assume a typical pattern of flux variation** - From initial time \( t=0 \) up to \( t=8\,s \): flux **increases linearly** from 0 to a maximum value \( \Phi_{max} \). - From \( t=8\,s \) to \( t=12\,s \): flux remains **constant** at \( \Phi_{max} \). - From \( t=12\,s \) to \( t=16\,s \): flux **decreases linearly** from \( \Phi_{max} \) back to 0. - From \( t=16\,s \) onward: flux remains **constant** at 0. **Step 3: Determine the induced emf and current in each interval** - The induced emf is given by Faraday's Law: \[ \mathcal{E}(t) = - \frac{d\Phi_B(t)}{dt} \] - The induced current: \[ I(t) = \frac{\mathcal{E}(t)}{R} \] - Sign conventions: - When \( \frac{d\Phi_B}{dt} > 0 \), emf is negative, inducing a current **opposite** to the flux increase. - When \( \frac{d\Phi_B}{dt} < 0 \), emf is positive, inducing a current **to support** the original flux. **Step 4: Calculate the emf and current during each interval** - **0 to 8 seconds (flux increasing):** \[ \frac{d\Phi_B}{dt} = \frac{\Phi_{max}}{8} \] \[ \mathcal{E} = - \frac{\Phi_{max}}{8} \] \[ I(t) = - \frac{\Phi_{max}}{8 R} \] - The negative emf indicates the current flows **opposite** to the assumed positive direction (say, CCW), meaning **CCW current**. - **8 to 12 seconds (flux constant):** \[ \frac{d\Phi_B}{dt} = 0 \Rightarrow \mathcal{E} = 0 \Rightarrow I(t) = 0 \] - **12 to 16 seconds (flux decreasing):** \[ \frac{d\Phi_B}{dt} = - \frac{\Phi_{max}}{4} \quad (\text{since decreasing over 4 seconds}) \] \[ \mathcal{E} = - \left(- \frac{\Phi_{max}}{4}\right) = + \frac{\Phi_{max}}{4} \] \[ I(t) = \frac{\mathcal{E}}{R} = + \frac{\Phi_{max}}{4 R} \] - The positive emf causes the current to **support** the original flux (e.g., if initially CCW, now CW). - **16 to 18 seconds (flux constant at zero):** \[ I(t) = 0 \] **Step 5: Sketch the current graph** - From 0 to 8 s: current is **constant and positive** \( \displaystyle - \frac{\Phi_{max}}{8 R} \). - From 8 to 12 s: current drops to **zero**. - From 12 to 16 s: current jumps to **positive** \( \displaystyle \frac{\Phi_{max}}{4 R} \). - From 16 to 18 s: back to **zero**. --- ### **Part B: Justification of Current Direction between 12–16 Seconds** - During this interval, the flux **decreases**. - According to **Lenz's Law**, the induced current **flows in a direction** to oppose the decrease. - If initially the flux was increasing CCW, then decreasing flux implies the induced current **flows CW** to support the original flux. - Therefore, **the induced current flows CW** during 12–16 seconds. --- ### **Part C: Energy Dissipated in the Loop during the First 8 Seconds** **Step 1: Determine the emf during flux increase** - Flux increases linearly from 0 to \( \Phi_{max} \) over 8 seconds. \[ \frac{d\Phi_B}{dt} = \frac{\Phi_{max}}{8} \] - Induced emf: \[ \mathcal{E} = - \frac{\Phi_{max}}{8} \] **Step 2: Determine the induced current** \[ I = \frac{\mathcal{E}}{R} = - \frac{\Phi_{max}}{8 R} \] - The magnitude of current: \[ |I| = \frac{\Phi_{max}}{8 R} \] **Step 3: Calculate the power dissipated** - Power at any instant: \[ P = I^2 R = \left(\frac{\Phi_{max}}{8 R}\right)^2 R = \frac{\Phi_{max}^2}{64 R} \] - Since the emf and current are constant during 0–8 s, total energy dissipated: \[ E = P \times t = \frac{\Phi_{max}^2}{64 R} \times 8 = \frac{\Phi_{max}^2}{8 R} \] --- ### **### **Final Results:** - **Induced current \( I(t) \):** - **0–8 s:** \( I(t) = - \frac{\Phi_{max}}{8 R} \) (CCW) - **8–12 s:** \( I(t) = 0 \) - **12–16 s:** \( I(t) = + \frac{\Phi_{max}}{4 R} \) (CW) - **16–18 s:** \( I(t) = 0 \) - **Direction justification:** During decreasing flux, the induced current flows CW to oppose the flux decrease. - **Total energy dissipated during first 8 seconds:** \[ \boxed{ E = \frac{\Phi_{max}^2}{8 R} } \] --- **Note:** To get a numerical value, insert actual values of \( \Phi_{max} \) and \( R \). If you provide them, I can help with the computation!