Let's carefully analyze and solve the problem step-by-step. --- ### Given Data: - Currents: - \( I_1 = 3.16\, \text{A} \) - \( I_2 = 4.65\, \text{A} \) - Distance between the wires: - \( d = 10.0\, \text{cm} = 0.10\, \text{m} \) - The wires are long and parallel, with currents flowing in the indicated directions (assumed to be specified, but since it's not explicitly given, we'll analyze the general case: assume they are either both in the same direction or opposite). --- ### **Part (a): Magnetic field at the midpoint between the wires** --- #### Step 1: Identify the position - The point is **midway** between the two wires, so: - Distance from each wire to the midpoint: \[ r = \frac{d}{2} = \frac{0.10\, \text{m}}{2} = 0.05\, \text{m} \] --- #### Step 2: Magnetic field due to each wire (Biot–Savart Law / Ampère's Law) The magnetic field at a distance \( r \) from a long straight current-carrying wire: \[ B = \frac{\mu_0 I}{2 \pi r} \] where \[ \mu_0 = 4\pi \times 10^{-7} \,\text{T}\cdot\text{m/A} \] --- #### Step 3: Calculate each magnetic field magnitude - For wire 1: \[ B_1 = \frac{\mu_0 I_1}{2 \pi r} = \frac{(4\pi \times 10^{-7}) \times 3.16}{2 \pi \times 0.05} \] Simplify numerator and denominator: \[ B_1 = \frac{4\pi \times 10^{-7} \times 3.16}{2 \pi \times 0.05} = \frac{(4 \times 10^{-7}) \times 3.16}{2 \times 0.05} \] \[ B_1 = \frac{(1.264 \times 10^{-6})}{0.10} = 1.264 \times 10^{-5}\,\text{T} \] - For wire 2: \[ B_2 = \frac{\mu_0 I_2}{2 \pi r} = \frac{(4\pi \times 10^{-7}) \times 4.65}{2 \pi \times 0.05} \] Similarly: \[ B_2 = \frac{4 \times 10^{-7} \times 4.65}{0.10} = \frac{1.86 \times 10^{-6}}{0.10} = 1.86 \times 10^{-5}\,\text{T} \] --- #### Step 4: Determine directions of the magnetic fields - The magnetic field due to a current-carrying wire is tangent to a circle around the wire, following the right-hand rule: - If the current is **upward** (say), the magnetic field encircles the wire in a **counterclockwise** direction when looking from above. - **Assumption:** Both currents are in the same direction (say, both upward). Alternatively, if they are in opposite directions, the field directions will change accordingly. **Since the problem states "currents of \( I_1 \) and \( I_2 \) in the indicated directions," and the directions are not specified explicitly,** let's assume **both currents are in the same direction** (upward). - At the midpoint: - The magnetic field from wire 1 will be **circulating** around wire 1, with the field at the midpoint pointing **perpendicular** to the line connecting wire 1 and the midpoint, following the right-hand rule. - The same applies for wire 2. **At the midpoint**: - Both magnetic fields will have **tangential directions**, which will be **perpendicular to the line connecting the wires**. - By applying the right-hand rule: - For an upward current, the magnetic field at a point **to the right** of the wire points **out of the page**, and **to the left** of the wire points **into the page**. Since the point is midway: - The magnetic field from wire 1 will point **out of the page** on one side. - The magnetic field from wire 2 will also point **out of the page** if the currents are in the same direction. **Thus, the magnetic fields due to both wires at the midpoint are in the same direction (out of the page).** --- #### Step 5: Combine the magnetic fields Total magnetic field at the midpoint: \[ B_{\text{total}} = B_1 + B_2 = 1.264 \times 10^{-5} + 1.86 \times 10^{-5} = 3.124 \times 10^{-5}\,\text{T} \] --- ### **Final answer for Part (a):** \[ \boxed{ \textbf{Magnitude:} \quad B = 3.124 \times 10^{-5}\, \text{T} } \] \[ \boxed{ \textbf{Direction:} \quad \textbf{out of the page} } \] --- ### **Part (b): Magnetic field at point \( P \), located 10.0 cm above wire 1** --- #### Step 1: Position of point \( P \) - \( P \) is located **above wire 1** at a distance \( d = 10.0\, \text{cm} = 0.10\, \text{m} \). - The position relative to wire 1: - Distance from wire 1: \( r_1 = 0.10\, \text{m} \) - Position relative to wire 2: - The wires are separated by 0.10 m, so the distance from wire 2 to \( P \): \[ r_2 = \sqrt{d^2 + \left(\frac{d}{2}\right)^2} = \sqrt{(0.10)^2 + (0.05)^2} = \sqrt{0.01 + 0.0025} = \sqrt{0.0125} \approx 0.1118\, \text{m} \] --- #### Step 2: Magnetic field due to wire 1 at \( P \): \[ B_1' = \frac{\mu_0 I_1}{2 \pi r_1} = \frac{4\pi \times 10^{-7} \times 3.16}{2 \pi \times 0.10} = \frac{(4 \times 10^{-7}) \times 3.16}{0.20} = \frac{1.264 \times 10^{-6}}{0.20} = 6.32 \times 10^{-6}\,\text{T} \] - **Direction:** Using the right-hand rule: - Current upward, point directly **above** the wire: - The magnetic field circles around the wire. - At a point **above** the wire, the magnetic field due to wire 1 points **counterclockwise** when viewed from above. - For a point **above** the wire (on the same side as the magnetic field circulation), the magnetic field points **into the page** (assuming upward current). --- #### Step 3: Magnetic field due to wire 2 at \( P \): \[ B_2' = \frac{\mu_0 I_2}{2 \pi r_2} = \frac{4\pi \times 10^{-7} \times 4.65}{2 \pi \times 0.1118} = \frac{(4 \times 10^{-7}) \times 4.65}{2 \times 0.1118} \] Calculate numerator: \[ 4 \times 10^{-7} \times 4.65 = 1.86 \times 10^{-6} \] Calculate denominator: \[ 2 \times 0.1118 = 0.2236 \] Thus: \[ B_2' = \frac{1.86 \times 10^{-6}}{0.2236} \approx 8.32 \times 10^{-6}\, \text{T} \] - **Direction:** With the same assumptions: - The magnetic field due to wire 2 at \( P \) (which is **above** wire 2), will **also** point **into the page** if currents are both upward, because of the right-hand rule. --- #### Step 4: Vector addition of the fields - Both magnetic fields point **into the page**: \[ B_{\text{total}} = B_1' + B_2' = 6.32 \times 10^{-6} + 8.32 \times 10^{-6} = 1.464 \times 10^{-5}\, \text{T} \] --- ### **Final answer for Part (b):** \[ \boxed{ \textbf{Magnitude:} \quad B \approx 1.464 \times 10^{-5}\, \text{T} } \] \[ \boxed{ \textbf{Direction:} \quad \textbf{into the page} } \] --- ### **Summary:** | Part | Magnitude of Magnetic Field | Direction | |--------|------------------------------|------------------------------| | (a) | \( 3.124 \times 10^{-5}\, \text{T} \) | Out of the page | | (b) | \( 1.464 \times 10^{-5}\, \text{T} \) | Into the page | --- **Note:** The directions depend on the actual current directions given in the problem; the above assumes both currents are upward. If currents are in opposite directions, the field directions would change accordingly.