Moving Charges & Magnetism Notes PDF

# Moving Charges & Magnetism ## Day 1 - Any conductor carrying current produces magnetic field around it. - S.I Unit for B is Tesla (T). - **Oersted's Law** - A current carrying conductor produces a magnetic field around it (circular path). - **Maxwell's Right Hand Thumb Rule** - **Thumb:*...

# Moving Charges & Magnetism ## Day 1 - Any conductor carrying current produces magnetic field around it. - S.I Unit for B is Tesla (T). - **Oersted's Law** - A current carrying conductor produces a magnetic field around it (circular path). - **Maxwell's Right Hand Thumb Rule** - **Thumb:** Current (I) - **Folded fingers:** Magnetic field (B) - **Screw Rule** - Current flowing **inwards:** Magnetic field **clockwise** - Current flowing **outwards:** Magnetic field **anticlockwise** - **Circular Coil** - **Right Hand Thumb Rule:** - **Thumb:** Magnetic (B) - **Folded fingers:** Current (I) - **Screw Rule** - Current flowing **inwards:** Magnetic field **towards observer** - Current flowing **outwards:** Magnetic field **away from observer** - **Measurement of Magnetic Field (B)** - **Biot-Savart Law** - The magnitude of magnetic field element ($dB$) due to a small current element (Idl) is: - $dB = \frac{μ_o}{4π} \frac{Idl sin θ}{r^2}$, - where θ is the angle between the current element and the line joining the current element and the point P. - $μ_o$ is called as permeability of free space: - $μ_o = 4π × 10^{-7} $ - **Magnetic field produced around a wire** - If the current element points perpendicular to the radius vector: $dB = \frac{μ_o}{4π}\frac{Idl sin 90°}{r^2}$. - If the current element points along the radius vector: $dB = 0$. - **Expression for Biot-Savart law in vector form:** - $dB = \frac{μ_o I}{4π}\frac{(dl × \overrightarrow{r})}{r^3}$ - **Proof:** - $dB = \frac{μ_oI}{4π} \frac{(dl × \overrightarrow{r})}{r^3}$ - $dB = \frac{μ_oI}{4π}\frac{dl sin θ}{r^2} × \overrightarrow{r}$ - $\overrightarrow{a} × \overrightarrow{b}= ab sin θ$ - $dB = \frac{μ_o I}{4π}\frac{dl sin θ}{r^2} × \overrightarrow{r}$ ## Consider a Circular current - Apply Biot Savart law: - $dB = \frac{μ_o Idl sin θ}{4πr^2}$ - $dB = \frac{μ_o Idl sin 90°}{4πr^2}$ - $dB = \frac{μ_o Idl}{4πr^2}$ - The direction of the magnetic field at the centre is the same. - $B_c = \int db= \int \frac{μ_oIdl}{4πr^2} = \frac{μ_oI}{4πr^2} \int dl$ - $B_c = \frac{μ_oI}{4πr^2} × 2πr$ - $B_c = \frac{μ_oI}{2r}$ - **For N turns:** - $B_c = \frac{μ_oNI}{2r}$ ## A circular coil of wire consisting of 100 turns, each of radius - $B_c = \frac{μ_oNI}{2r}$ - $B_c = \frac{4π × 10^{-7} × 100 × 0.40}{2 × 8 × 10^{-2}}$ - $B_c = 3.14 × 10^{-4}$ T ## Example 4.5 An element ($Δl = Δx$) is placed at the origin and carries - Apply biot-savart Law: - $dB = \frac{μ_oIdl sin θ}{4πr^2}$ - $dB = \frac{μ_oIdl sin 90°}{4πr^2}$ - $dB = \frac{4π × 10^{-7} × 10 × 1 × 10^{-2} × 1}{4π × (0.5)^2 }$ - $dB = 4 × 10^{-8}$ T - direction of dB is outwards from Maxwell's right hand thumb rule. ## Example 4.6 A straight wire carrying a current of 12 A is bent into a semi-circular arc. - **(a) Magnetic field due to the straight segments:** - Apply Biot-Savart Law: - $dB = \frac{μ_oIdl sin θ}{4πr^2}$ - $dB = 0$ - **(b) Circular coil:** - $B_c = \frac{μ_oI}{2r}$ - **Semicircular arc:** - $B_c = \frac{μ_oI}{4r}$ - **(c) From Fig. 1:** - $B_1 = \frac{μ_oI}{4r}$ (Inwards) - $B_1 = \frac{4π × 10^{-7} × 12}{4 × 2×10^{-2}}$ - $B_1 = 6π × 10^{-5}$ T (Inwards) - **From Fig 2:** - $B_2 = \frac{μ_oI}{4r}$ (Outwards) - $B_2 = \frac{4π × 10^{-7} × 12}{4 × 2×10^{-2}}$ - $B_2 = 6π × 10^{-5}$ T (Outwards) - $B_1 = B_2$, but the directions of $B_1$ and $B_2$ are different. ## Day 3 - **Two identical coils P and Q, each of radius R** - $B_c = \frac{μ_oI}{2r}$ - $B_p = \frac{μ_oI}{2r}$ - $B_q = \frac{μ_o√3 I}{2r}$ - $B_q = √3 B_p$ - **Find the magnitude of the net magnetic field at the common centre of the two coils:** - $B_{net}^2 = B_p^2 + B_q^2$ - $B_{net}^2 = B_p^2 + 3B_p^2$ - $Β_{net} = 2B_p = 2 × \frac{μ_oI}{2r}$ - $Β_{net} = \frac{μ_oI}{r}$ - $Β{net} = 30°$ with $B_q$ ## Example 4.14 Two concentric circular coils X and Y of radii - **For Coil X:** - $B_x = \frac{μ_oNI}{2r}$ - $B_x = \frac{4π × 10^{-7} × 20 × 16}{2 × 16 × 10^{-2}}$ - $B_x = 4π × 10^{-4} $ T (towards East) - **For Coil Y:** - $B_y = \frac{μ_oNI}{2r}$ - $B_y = \frac{4π × 10^{-7} × 25 × 18}{2 × 10 × 10^{-2}}$ - $B_y = 9π × 10^{-4}$ T (towards west) - $B_y > B_x$ - **Net Magnetic field:** - $B_{net} = B_y - B_x$ - $Β_{net} = 9π × 10^{-4} - 4π × 10^{-4}$ - $Β_{net} = 5π × 10^{-4}$ T (towards west) ## How is the magnetic field intensity at the centre of a circular coil carrying a current 'I'? - $B = \frac{μ_oI}{2r}$ - If current (I) is doubled: $I' = 2I$ - If radius (r) is made half: $r' = \frac{r}{2}$ - $B' = \frac{μ_oI'}{2r'}$ - $B' = \frac{μ_o × 2I}{2 × \frac{r}{2}}$ - $B' = 4B$ ## A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having 'n' turns. - **Case (i)** - $N_1 = 1$, $r_1 = 2πr_1$, $I_1 = 1 × 2πr$ - $B_1 = \frac{μ_oN_1 I_1}{2r_1}$ - $B_1 = \frac{μ_o × 1 × 2πr}{2 × 2πr}$ - $B_1 = \frac{μ_oI}{2}$ - **Case (ii)** - $N_2 = n$, $r_2 = 2πr_2$, $I_2 = 1× 2πr_2$ - $B_2 = \frac{μ_oN_2I_2}{2r_2}$ - $B_2 = \frac{μ_o × n × 2πr}{2 × 2πr}$ - $B_2 = \frac{μ_onI}{2}$ - **Ratio of magnetic fields:** - $\frac{B_2}{B_1} = \frac{μ_onI}{2} × \frac{2}{μ_oI}$ - $\frac{B_2}{B_1} = n^2$ - $B_1 : B_2 = 1 : n^2: $ ## Two circular coils X and Y having radii R and R/2 respectively in a horizontal plane with their centres coinciding with each other. - **At point O:** - $B_x = B_y$ (opposite direction) - $B_x = \frac{μ_oI}{2R}$ - $B_y = \frac{μ_oI'}{2 × \frac{R}{2}}$ - $I' = 2I$ ## Find an expression for magnetic field due to a circular coil at a distance x on the axial line. - Apply Biot Savart Law: - $dB = \frac{μ_oIdl sin θ}{4πR^2}$ - $dB = \frac{μ_oIdl sin 90°}{4πR^2}$ - $dB = \frac{μ_oIdl}{4πR^2}$ - $R = (r^2 + x^2)^{1/2}$ - **Resolve dB into 2 components:** - $dB sin θ$ - $dB cos θ$ - The vertical components ($dB cos θ$) cancel each other. - The horizontal components ($dB sin θ$) add up. - $B_{axial} = \int dB sin θ$ - $B_{axial} = \int \frac{μ_oIdl}{4πR^2} × \frac{r}{R}$ - $B_{axial} = \frac{μ_oIr}{4πR^3} \int dl = \frac{μ_oIr}{4πR^3} × 2πr$ - $B_{axial} = \frac{μ_oIr^2}{2R^3}$ - $B_{axial} = \frac{μ_oIr^2}{2(r^2 + x^2)^{3/2}}$ - **For N turns:** - $B_{axial} = \frac{μ_oNIr^2}{2(r^2 + x^2)^{3/2}}$ ## Two small identical circular loops, marked (1) and (2), carrying equal currents, are placed with the geometrical axes perpendicular to each other as shown in the figure. - **Apply Maxwell's Right Hand Thumb Rule:** - $B_1 = \frac{μ_oIr^2}{2(r^2 + x^2)^{3/2}}$ - $B_2 = \frac{μ_oIr^2}{2(r^2 + x^2)^{3/2}}$ - **Find the magnitude and direction of the net magnetic field produced at the Point O:** - $B_{net}^2 = B_1^2 +B_2^2$ - $B_{net} = √B_1^2 +B_2^2 = √2B_1$ - $B_{net} = √2\frac{μ_oIr^2}{2(r^2 + x^2)^{3/2}}$ - $B_{net} = \frac{μ_oIr^2 √2}{2(r^2 + x^2)^{3/2}}$ - $θ = 45°$ with $B_1$ ## Day 4 - **Special Case:** - **At Centre (x = 0):** - $B_{centre} = \frac{μ_oNIr^2}{2r^3} $ - $B_{centre} = \frac{μ_oNI}{2r}$ - $B_{centre} = \frac{μ_oNI}{2r}$ ## Two identical circular loops, P and Q, each of radius r and carrying currents I and 2I respectively are lying in parallel planes such that they have a common axis. - **Find the magnitude of the net magnetic field at point O.** - $B_{axial} = \frac{μ_oIr^2}{2(r^2+x^2)^{3/2}}$ - $B_p = \frac{μ_oIr^2}{2(r^2 + r^2)^{3/2}}$ - $B_p = \frac{μ_oIr^2}{2(2r^2)^{3/2}}$ - $B_q = \frac{μ_o(2I)r^2}{2(r^2 + r^2)^{3/2}}$ - $B_q = \frac{μ_o(2I)r^2}{2(2r^2)^{3/2}}$ - $B_q = 2B_p$ - $Β_{net} = B_q - B_p$ - $Β{net}= 2B_p - B_p$ - $Β{net} = B_p = \frac{μ_oIr^2}{2(2r^2)^{3/2}}$ - $Β{net}= \frac{μ_oI}{4√2r}$ (towards Q) ## Two Parallel coaxial circular coils - **Apply Biot Savart Law:** - $Β_{axial} = \frac{μ_oNIr^2}{2(r^2+x^2)^{3/2}}$ - $B_1 = \frac{μ_oNIR^2}{2(R^2 + R^2)^{3/2}}$ - $B_2 = \frac{μ_oNIR^2}{2(R^2 + R^2)^{3/2}}$ - **Net magnetic field at the midpoint of the line joining their centres:** - $B_{net} = B_1 + B_2$ - $B_{net} = 2B_1 = 2 × \frac{μ_oNIR^2}{2(2R^2)^{3/2}}$ - $B_{net} = \frac{μ_oNIR^2}{2√2R^3}$ - $B_{net} = \frac{μ_oNI}{2√2R}$ (towards 1) ## Ampere's Circuital Law - This law gives a relation between I, s, and B. - This law is similar to Gauss's Law (used to calculate electric field). - **Formula:** - $\oint B.dl = μ_oI$ - Where I is the total current enclosed by the closed path. - **Proof** - **Total Current:** - $ I_{total} = (I_2 + I_3 - I_4 - I_5 + I_6)$ - **Applying Ampere's Circuital Law:** - $\oint B . dl = μ_oI_{total}$ - $\int_{AB} B.dl + \int_{BC} B.dl + \int_{CD} B.dl + \int_{DA} B.dl = μ_oI_{total}$ - $B \int_{AB} dl = μ_oI_{total}$ ($B = 0$ on BC, CD, S4 DA) - $B × l = μ_oI_{total}$ - $B = \frac{μ_oI_{total}}{l}$ - Note: The Ampere's surface can be of any shape but for calculations, we try to take symmetric surfaces where the angle between B and dl will be 0 and it will be easy to calculate. ## Find an expression for magnetic field due to an infinite long current carrying conductor. - **Apply Ampere's Circuital Law:** - $\oint B.dl = μ_oI$ - $B \int dl = μ_oI$ - $B (2πr) = μ_oI$ - $B = \frac{μ_oI}{2πr}$ ## Day 6 - **A long straight wire carries a current of 35 A** - $B = \frac{μ_oI}{2πr}$ - $B = \frac{4π × 10^{-7}× 35}{2π × 20 × 10^{-2}}$ - $B =35 × 10^{-6}$ T ## Example 4.8 Figure 4.15 shows a long straight wire of a circular cross-section (radius a) carrying steady current I. - **Case (i) r < a** - Apply Ampere's Circuital Law: - $\oint B . dl = μ_oI$ - $\oint Bdl cos θ = \oint Bdl$ - $B \oint dl = B (2πr)$ - **Find I enclosed:** - $I_e = \frac{Area (A')}{Area (A)} × I = (\frac{πr^2}{πa^2}) × I$ - $I_e = \frac{I r^2}{a^2}$ - **Applying Ampere's Circuital Law:** - $B(2πr) = μ_o\frac{I r^2}{a^2}$ - $B_{inside} = \frac{μ_oIr}{2πa^2}$ - **Case (ii) r > a** - Apply Ampere's Circuital Law: - $\oint B . dl = μ_oI$ - $\oint Bdl cos θ = \oint Bdl$ - $B \oint dl = B (2πr)$ - **Find I enclosed:** - $I_e = I$ - **Applying Ampere's Circuital Law:** - $B(2πr) = μ_oI$ - $B_{outside} = \frac{μ_oI}{2πr}$ ## Day 7 - **Consider a long current carrying solenoid:** - Apply Ampere's Circuital Law: - $\oint B.dl = μ_oI$ - **For L.H.S:** - $\oint B.dl = \int_{AB} Bdl + \int_{BC} Bdl + \int_{CD} Bdl + \int_{DA}Bdl$ - $\oint B.dl = B \int_{AB} dl + B \int_{BC} dl + B \int_{CD} dl + B \int_{DA}Bdl$ - $\oint B.dl = B \int_{AB} dl = Bl$ - **For R.H.S** - $I_e = NI_l$ - $I_e = nLl$ - **Equating L.H.S and R.H.S:** - $Bl = μ_o nLI$ - $B = μ_onI$ ## The magnetic field at the end of a solenoid - $B_{end} = \frac{1}{2}μ_onI$ ## Toroidal Solenoid or Toroid - A toroid has no open ends; it forms a closed loop. - **Inside the core:** - $B.dl = μ_oI$ - $B × 2πr = μ_oI$ - $B_{inside} =\frac{μ_oI}{2πr} $ - **Outside the core:** - $B. dl = μ_oI$ - $I_e = 0$ - $B$ is zero. - **Qualitiative Study:** - $Β_{inside}= μ_onI$ ## Day 8 - **Consider a current carrying toroidal solenoid, having 'n' number of turns per unit length:** - **For inside:** - Apply Ampere's Circuital Law -$\oint B. dl =μ_oI$ - $\oint Bdlcos θ = \oint Bdl$ - $B\oint dl = B(2πr)$ - **Find I Enclosed:** - $I_e = N(2πr)I$ - $I_e = n(2πr)l$ - **Equating L.H.S and R.H.S** - $B(2πr) = μ_on(2πr)l$ - $B_{core} = μ_onI$ - **For Outside:** - Apply Ampere's Circuital Law - $\oint B.dl = μ_oI$ - $I_e = 0$ - $B_{outside} = 0$ - **Example 4.9 A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?** - $l = 0.5m$, $r = 1 cm$, $N = 500$, $I = 5A$ - $n = \frac{N}{l} = \frac{500}{0.5} = 1000$ turns per cm - $B_{core} = μ_onI = 4π × 10^{-7} × 1000 × 5 = 6.28 × 10^{-3}$ T - **Example 4. 17 A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid?** - **(a) Outside the toroid:** - $B_{outside} = 0$ - **(b) Inside the core of the toroid:** - $B_{core} = μ_onI$ - $n = \frac{N}{2πr} = \frac{3500}{2π × 25.5 × 10^{-2}}$ - $B_{core} = 4π × 10^{-7}× \frac{3500}{2π × 25.5 × 10^{-2}} × 11 ≈ 3.02 × 10^{-2}$ T - **(c) In the empty space surrounded by the toroid:** - $B_{empty} = 0$ ## Force experienced by a charge moving in a magnetic field B - **Electric force:** - $F_E = qE$ - If the charge is at rest, the force is only due to the electric field. - This force increases velocity and accelerates the charge. - Direction: Along the direction of the electric field. - **Magnetic force:** - $F_B = qvB sin θ$ - At an angle (θ = 0° or 180°), $F_B = 0$. - This force changes the direction of the charge but does not increase or decrease its velocity. - Direction: Perpendicular to both velocity and magnetic field. - **Lorentz force:** - $F = F_E + F_B = qE + q[v × B]$ - This is the total force acting on a charged particle moving in both electric and magnetic fields. ## To find the direction of force - Apply Fleming's Left Hand Rule: - **Index finger:** Magnetic field (B) - **Middle finger:** Velocity (v) - **Thumb:** Force (F) ## Time period of the charge particle - $T= \frac{2πm}{qB}$ - The time period does not depend on r. ## Why is the charge does not experience an increase in K.E when it enters normally into the magnetic field? Workdone (W) is required to increase K.E. - $W = F × s × cos θ$ - If the angle between Force and displacement is 90°: $cos θ = 0$, so W =0. ## Example 4. 2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge). - **(a) Electron:** - $F= [v × B]$ - $F = -e[v × B(-j)]$ - $F = +evB[i × j]$ - $F = eVB[k]$ - **(b) Proton:** - $F= [v × B]$ - $F = +e[v × B(-j)]$ - $F = +evB[i × j]$ - $F = +eVB[k]$ ## Velocity Selector or Velocity Filter - This technique selects a particular velocity from a beam of particles moving in a crossed field (electric and magnetic). - When a charged particle is moving in a crossed field, it will experience both electric (F_E) and magnetic (F_B) forces. - The velocity (v) of the particle which goes undeflected is given by: - v = E/B ## Case (i) If v < v_o: - $F_B < F_E$ - The particle will move upwards and hit the wall. - $F_B > F_E$ - The particle will move downwards and hit the wall. - If $v > v_o$: - $F_B > F_E$ - The particle will move upwards and hit the wall. - $F_B < F_E$ - The particle will move downwards and hit the wall. - If $v = v_o$: - $F_B = F_E$ - The particle will go undeflected. ## Motion of a charged particle in a magnetic field at an angle θ (Helical motion) - **If θ = 0° or 180°** - $F_B = 0$ - The particle will go undeflected and it will continue with its path. - **If θ = 90°** - $F_{max} = qvB$ - The particle will undergo circular motion. - **If θ ≠ 0°, 90°** - The particle will undergo helical motion (a combination of circular and linear motion). ## Lorentz magnetic force - $F_B = qvB sin θ$ - $F_B = q[v × B]$ - This is the total force acting on a charged particle moving in a magnetic field. ## Day 11 - **Conversion of Galvanometer into Ammeter/ Voltmeter** - **Galvanometer:** Used to measure small current values (mA). - **Ammeter:** Used to measure a larger current values (Amperes). - **Voltmeter:** Used to measure the voltage across two points in a circuit. - **Conversion of Galvanometer into Ammeter** - **Principle:** A galvanometer is converted into an ammeter by connecting a low resistance (called shunt resistance, $r_s$) in parallel with it. - **Formula:** - $r_s = \frac{I_gR_g}{I_{max}-I_g}$ - **Conversion of Galvanometer into Voltmeter** - **Principle:** A galvanometer is converted into a voltmeter by connecting a high resistance (R) in series with it. - **Formula:** - $R = \frac{V_{max} - I_gR_g}{I_g}$ ## Day 12 - **A galvanometer coil has a resistance of 12 Ω and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?** - $R_g = 12$ ohms, $I_g = 3mA = 3 × 10^{-3}$ A, $V_{max} = 18V$ - $R = \frac{V_{max} - I_gR_g}{I_g}$ - $R = \frac{18 - 3×10^{-3}×12}{3 × 10^{-3}}$ - $R = 5988$ ohms - Answer: By connecting a resistance of 5988 ohms in series with the galvanometer, we can get a voltmeter of range 0 to 18V. - **A galvanometer has a resistance of 15 ohms and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6A?** - $R_g = 15$ ohms, $I_g = 4 mA = 4 × 10^{-3}$ A, $I_{max} = 6A$ - $r_s = \frac{I_gR_g}{I_{max}-I_g}$ - $r_s = \frac{4×10^{-3}×15}{6 - 4×10^{-3}}$ - $r_s = 0.01$ ohms - Answer: By connecting a resistance of 0.01 ohms in parallel with the galvanometer, we can get an ammeter of range 0 to 6A. - **In a galvanometer, there is a deflection of 10 divisions per mA. The galvanometer has 50 divisions per mA. If a shunt of 25 ohms is connected to the galvanometer and there are 50 divisions in all of the scale of the galvanometer. What maximum current can this galvanometer read?** - 10 divisions = 1 mA - 50 divisions = 5 mA - It means that the current sensitivity of the galvanometer is 10 divisions per mA. - $r_s = 25$ ohms, $I_g = 5 × 10^{-3}$ A - $I_{max} = \frac{I_gR_g + I_gr_s}{r_s}$ - $I_{max} = \frac{5 × 10^{-3} × 60 + 5 × 10^{-3} × 25}{25}$ - $I_{max} = 17 × 10^{-3}$ A = 17 mA - **A voltmeter of a certain range is constructed by connecting a resistance of 980 ohms in series with a galvanometer. When the resistance of 470 ohms is connected in series, the range gets halved. Find the resistance of the galvanometer.** - **Case 1:** - $R_1 = R_g + 980$ - $I_g(R_1 + R_g) = V_{max}$ - **Case 2:** - $R_2 = R_g + 470$ - $I_g(R_2 + R_g) = \frac{V_{max}}{2}$ - Dividing Case 1 by Case 2: - $\frac{R_g + 980}{R_g + 470} = 2$ - $R_g + 980 = 2R_g + 940$ - $R_g = 40$ ohms - **A multirange voltmeter can be constructed by using a galvanometer circuit as shown in the figure. We want to construct a voltmeter that can measure 2 V, 20 V and 200 V using a galvanometer of resistance 10 ohms and that produces maximum deflection for a current of 1 mA. Find the value of R1, R2 and R3 that have to be used.** - **Case (i):** - $R_1 = \frac{V_{max} - I_gR_g}{I_g}$ - $R_1 = \frac{2 - 1 ×10^{-3} × 10}{1 × 10^{-3}} = 1990$ ohms (R_1 = 2kΩ) - **Case (ii):** - $R_1 + R_2 = \frac{V_{max} - I_gR_g}{I_g}$ - $R_1 + R_2 = \frac{20 - 1 × 10^{-3} × 10}{1 × 10^{-3}} = 19990$ ohms (R_1 + R_2 ≈ 20kΩ) - $R_2 ≈ 18,000$ ohms (R_2 ≈ 18 kΩ) - ** Case (iii):** - $R_1