Let's solve it **step by step**: --- ### **Given Data** - Current, \( I = 5.00\,\text{A} \) - Loop radius, \( R = 0.100\,\text{m} \) - Magnetic field at distance \( x \) from the center of the loop: \[ B(x) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] - At point A, \( x = R \) - Charge, \( q = -2.00\,\text{nC} = -2.00 \times 10^{-9}\,\text{C} \) - Velocity, \( \vec{v} = 375\,\frac{m}{s} \hat{k} \) --- ## **(i) B field at Point A due to the loop** \[ B_{\text{loop}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] At Point A, \( x = R \): \[ B_{\text{loop}} = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} \] \[ (2R^2)^{3/2} = (2^{3/2})(R^2)^{3/2} = 2^{3/2} R^3 = 2\sqrt{2} R^3 \] So, \[ B_{\text{loop}} = \frac{\mu_0 I R^2}{2 \cdot 2\sqrt{2} R^3} = \frac{\mu_0 I}{4\sqrt{2} R} \] **Direction:** Using the right-hand rule, the field at Point A points **upward** (in \(+\hat{y}\)). --- ## **(ii) B field at Point A due to the long line** The magnetic field a distance \( r \) from a long wire is: \[ B_{\text{wire}} = \frac{\mu_0 I}{2\pi r} \] At Point A, the distance from the wire is \( 2R \): \[ B_{\text{wire}} = \frac{\mu_0 I}{2\pi (2R)} = \frac{\mu_0 I}{4\pi R} \] **Direction:** Using the right-hand rule, the field at Point A points **downward** (in \(-\hat{y}\)), since the current is into the page. --- ## **(iii) Force on the charge at Point A** Total B field at A: \[ \vec{B}_{\text{total}} = (B_{\text{loop}} - B_{\text{wire}}) \hat{y} \] The force on a moving charge: \[ \vec{F} = q \vec{v} \times \vec{B} \] Given \(\vec{v} = 375 \hat{k}\), \(\vec{B}_{\text{total}} = (B_{\text{loop}} - B_{\text{wire}})\hat{y}\): \[ \hat{k} \times \hat{y} = -\hat{x} \] So, \[ \vec{F} = q v (B_{\text{loop}} - B_{\text{wire}}) (-\hat{x}) \] Plug in numbers: \[ B_{\text{loop}} = \frac{\mu_0 I}{4\sqrt{2} R} \] \[ B_{\text{wire}} = \frac{\mu_0 I}{4\pi R} \] \[ B_{\text{loop}} - B_{\text{wire}} = \frac{\mu_0 I}{4 R} \left(\frac{1}{\sqrt{2}} - \frac{1}{\pi}\right) \] Plug in values: \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A} \) \( I = 5 \, \text{A} \) \( R = 0.1 \, \text{m} \) \[ B_{\text{loop}} - B_{\text{wire}} = \frac{4\pi \times 10^{-7} \cdot 5}{4 \cdot 0.1} \left(\frac{1}{\sqrt{2}} - \frac{1}{\pi}\right) \] \[ = \frac{4\pi \times 10^{-7} \cdot 5}{0.4} \left(\frac{1}{\sqrt{2}} - \frac{1}{\pi}\right) \] \[ = \frac{20\pi \times 10^{-7}}{0.4} \left(\frac{1}{\sqrt{2}} - \frac{1}{\pi}\right) \] \[ = 50\pi \times 10^{-7} \left(\frac{1}{\sqrt{2}} - \frac{1}{\pi}\right) \] Calculate: \[ 50\pi \times 10^{-7} \approx 157 \times 10^{-7} = 1.57 \times 10^{-5} \] \[ \frac{1}{\sqrt{2}} \approx 0.707, \quad \frac{1}{\pi} \approx 0.318 \] \[ 0.707 - 0.318 = 0.389 \] \[ B_{\text{loop}} - B_{\text{wire}} \approx 1.57 \times 10^{-5} \cdot 0.389 \approx 6.1 \times 10^{-6} \, \text{T} \] Now, \( q = -2 \times 10^{-9} \), \( v = 375 \): \[ F = |q| v (B_{\text{loop}} - B_{\text{wire}}) \] \[ = 2 \times 10^{-9} \cdot 375 \cdot 6.1 \times 10^{-6} \] \[ = 750 \times 6.1 \times 10^{-15} \] \[ = 4575 \times 10^{-15} \] \[ = 4.6 \times 10^{-12} \, \text{N} \] **Direction:** \( -\hat{x} \), but since \( q \) is negative, force is in \( +\hat{x} \). **Final answer:** \( \boxed{4.6 \times 10^{-12} \hat{x} \, \text{N}} \) --- ## **(iv) Current in the long line for same B at A as loop** Set magnitudes equal: \[ B_{\text{loop}} = B_{\text{wire}} \] \[ \frac{\mu_0 I}{4\sqrt{2}R} = \frac{\mu_0 I'}{4\pi R} \] \[ \frac{I}{\sqrt{2}} = \frac{I'}{\pi} \] \[ I' = \frac{\pi}{\sqrt{2}} I \] \[ I' = \frac{3.142}{1.414} \times 5.00 \approx 2.22 \times 5.00 \approx 11.1 \, \text{A} \] --- ## **Summary of Final Answers** 1. **(i)** \( \boxed{\frac{\mu_0 I}{4\sqrt{2} R} \, \hat{y}} \) (upward) 2. **(ii)** \( \boxed{\frac{\mu_0 I}{4\pi R} \, (-\hat{y})} \) (downward) 3. **(iii)** \( \boxed{4.6 \times 10^{-12} \, \hat{x} \, \text{N}} \) 4. **(iv)** \( \boxed{I' = 11.1 \, \text{A}} \)