Let's break down the question step by step: --- ## **Given:** - **Current in wire (\(I\))**: 4.90 A (to the left) - **Distance from wire (\(r\))**: 15. cm = .150 m - **Charge (\(q\))**: \(+1.00 \, \mu C = 1.00 \times 10^{-6}\) C - **Velocity of charge (\(v\))**: \(2.00 \times 10^6\) m/s (to the right) --- ## **Step 1: Find the Magnetic Field (\(B\)) due to the Wire** The magnetic field at a distance \(r\) from a long straight wire is: \[ B = \frac{\mu_ I}{2\pi r} \] Where: - \(\mu_ = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\) Substitute the values: \[ B = \frac{4\pi \times 10^{-7} \times 4.90}{2\pi \times .150} \] \[ B = \frac{4 \times 10^{-7} \times 4.90}{.30} \] \[ B = \frac{19.6 \times 10^{-7}}{.30} \] \[ B = \frac{1.96 \times 10^{-6}}{.03} \] \[ B \approx 6.53 \times 10^{-6} \, \text{T} \] --- ## **Step 2: Find the Magnetic Force (\(F\))** The force on a moving charge in a magnetic field is: \[ F = q v B \sin\theta \] - The velocity of the charge is **perpendicular** to the magnetic field (since the wire is horizontal, the field circles the wire, and the charge moves parallel to the wire). Thus, \(\theta = 90^\circ\) and \(\sin\theta = 1\). \[ F = q v B \] Substitute the values: \[ F = (1.00 \times 10^{-6}) \times (2.00 \times 10^6) \times (6.53 \times 10^{-6}) \] \[ F = (2.00 \times 10^) \times (6.53 \times 10^{-6}) \] \[ F = 13.06 \times 10^{-6} = 1.31 \times 10^{-5} \, \text{N} \] --- ## **Step 3: Direction of the Force** - Use the **right-hand rule**: - Current is to the left. - Magnetic field at the location of the charge (above the wire) points **into the page** (using the right-hand grip rule). - The charge moves to the right. - Point your fingers to the right (velocity), curl them into the page (magnetic field); your thumb points **downward**. **Direction:** **Downward** (towards the wire). --- ## **Final Answers:** ### **Magnitude:** \[ \boxed{1.31 \times 10^{-5} \text{ N}} \] ### **Direction:** \[ \boxed{\text{Downward, towards the wire}} \]