Let's solve each part in **2 steps with explanations** and a final answer at the end of each part: --- ## (a) Sketch the Magnetic Field Lines and Describe the Field ### Step 1: Field Lines Inside the Toroid (\(a < r < b\)) - The current \(I\) flows around the toroid, creating circular magnetic field lines inside the "donut" (the core). - The field lines are **concentric circles** centered on the axis of the toroid, **tangent** to the path of the current. **Explanation:** By the right-hand rule, the magnetic field inside the toroid loops around the central axis, following the current direction. ### Step 2: Field Inside the “Donut Hole” (\(r < a\)) and Outside (\(r > b\)) - **Inside the hole (\(r < a\)):** No current is enclosed, so the field is **zero**. - **Outside the toroid (\(r > b\)):** Similarly, the net current enclosed is zero, so the field is **zero**. **Explanation:** Ampère’s law states that if no net current is enclosed, the magnetic field must be zero. **Final Answer:** - Inside the toroid (\(a < r < b\)): Circular, concentric field lines. - Inside the hole (\(r < a\)) and outside (\(r > b\)): Magnetic field is zero. --- ## (b) Symmetry of the Magnetic Field Inside ### Step 1: Shape and Direction - The magnetic field inside the toroid is **azimuthal** (circles around the central axis). **Explanation:** Due to the toroidal symmetry, the field must be tangent to circles centered on the axis. ### Step 2: Magnitude - The magnitude depends only on the distance \(r\) from the center, not on the angle. **Explanation:** By symmetry, at any point at radius \(r\), the field is the same. **Final Answer:** The field inside the toroid is **azimuthally symmetric** (depends only on distance from the center, not angle). --- ## (c) Ampère’s Law for Field Inside (\(a < r < b\)) ### Step 1: Apply Ampère’s Law \[ \oint \vec{B} \cdot d\vec{l} = \mu_ I_{\text{enc}} \] Choose a circular path of radius \(r\) (\(a < r < b\)), enclosing all \(N\) turns. ### Step 2: Calculate Enclosed Current \[ I_{\text{enc}} = N I \] \[ B \cdot 2\pi r = \mu_ N I \implies B = \frac{\mu_ N I}{2\pi r} \] **Explanation:** Ampère’s loop encloses all the current \(N I\), giving the result above. **Final Answer:** \[ B(r) = \frac{\mu_ N I}{2\pi r}, \quad a < r < b \] --- ## (d) Ampère’s Law for Field Outside (\(r < a\) and \(r > b\)) ### Step 1: For \(r < a\) (inside the hole) - The Ampère loop encloses **no current**. \[ I_{\text{enc}} = \implies B = \] ### Step 2: For \(r > b\) (outside the toroid) - The total net current passing through the loop is zero (current goes in and returns around the loop). \[ I_{\text{enc}} = \implies B = \] **Explanation:** By Ampère’s law, no net current enclosed means zero field. **Final Answer:** \[ B = \quad \text{for} \quad r < a \quad \text{and} \quad r > b \] --- ## (e) Is the Toroid Field Constant Like in a Solenoid? ### Step 1: Solenoid Field - In a long solenoid, the field inside is **constant**. ### Step 2: Toroid Field - In a toroid, \(B(r) \propto 1/r\), so it **varies** with radius. **Explanation:** Since \(r\) changes from \(a\) to \(b\), the field decreases as you move outward. **Final Answer:** The field inside a toroid is **not constant**; it decreases with increasing \(r\). --- ## (f) What if \(a\) and \(b\) Are Close Together? (Narrow Toroid) ### Step 1: Narrow Toroid (\(a \approx b\)) - The difference in \(r\) is small, so \(B(r)\) does not change much across the core. ### Step 2: Field Comparison - The field becomes **almost constant**, similar to a solenoid. - For \(r\) close to \(a\), \(B\) is slightly higher than for \(r\) close to \(b\). **Explanation:** As the width of the toroid shrinks, the \(1/r\) variation diminishes and the field approaches a constant value. **Final Answer:** If \(a \approx b\), the field inside the toroid becomes nearly constant, similar to a solenoid.