University:
Hunter College CUNY
Course:
PHYS 120.LB | Gen Elec-Magnetsm Lb
Academic year:

2023
Views:

370

Pages:

6
Author:

Simone Richards

Calculation of Magnetic Field at Point P Identify the Currents and Distances: There are two currents: 2 𝐴 2A and 5 𝐴 5A. Distances are 20 cm 20cm and 10 cm 10cm. Calculate Magnetic Fields: For the 2 𝐴 2A current: 𝐵 1 = 𝜇 0 ( 2 𝐴 ) 2 𝜋 ( 0.1 𝑚 ) = 4 𝜋 × 1 0 − 7 × 2 2 𝜋 × 0.1 = 4 × 1 0 − 6 T B 1 = 2π(0.1m) μ 0 (2A) = 2π×0.1 4π×10 −7 ×2 =4×10 −6 T For the 5 𝐴 5A current: 𝐵 2 = 𝜇 0 ( 5 𝐴 ) 2 𝜋 ( 0.1 𝑚 ) = 4 𝜋 × 1 0 − 7 × 5 2 𝜋 × 0.1 = 1 × 1 0 − 5 T B 2 = 2π(0.1m) μ 0 (5A) = 2π×0.1 4π×10 −7 ×5 =1×10 −5 T Vector Addition of Magnetic Fields: Since the fields are perpendicular: ∣ 𝐵 ⃗ ∣ = 𝐵 1 2 + 𝐵 2 2 = ( 4 × 1 0 − 6 ) 2 + ( 1 × 1 0 − 5 ) 2 ≈ 1.1 × 1 0 − 5 T ∣ B ∣= B 1 2 +B 2 2 = (4×10 −6 ) 2 +(1×10 −5 ) 2 ≈1.1×10 −5 T Electrical Power and Resistance Calculation Power Dissipation in a Resistor: Given 𝑉 = 120 𝑉 V=120V, 𝑃 = 800 𝑊 P=800W, and a resistor of 0.5 Ω 0.5Ω: Current calculation: 𝑃 appliance = 𝐼 × 𝑉 ⟹ 800 = 𝐼 × 120 ⟹ 𝐼 = 6.67 𝐴 P appliance =I×V⟹800=I×120⟹I=6.67A Power dissipated in the wire: 𝑃 dissipated = 𝐼 2 × 𝑅 = ( 6.67 𝐴 ) 2 × 0.5 Ω = 22 𝑊 P dissipated =I 2 ×R=(6.67A) 2 ×0.5Ω=22W Alternative Calculation: Find resistance of the appliance: 𝑅 = 𝑉 2 𝑃 = ( 120 ) 2 800 = 18 Ω R= P V 2 = 800 (120) 2 =18Ω Recalculate the current: 𝑉 = 𝐼 𝑅 ⟹ 120 = 𝐼 × ( 0.5 + 18 ) ⟹ 𝐼 = 6.49 𝐴 V=IR⟹120=I×(0.5+18)⟹I=6.49A Recalculate power dissipated in the wire: 𝑃 dissipated = 𝐼 2 × 𝑅 = ( 6.49 𝐴 ) 2 × 0.5 Ω = 21 𝑊 P dissipated =I 2 ×R=(6.49A) 2 ×0.5Ω=21W Q.2: Capacitors in Series Given: 𝐶 1 = 20 𝜇 𝐹 C 1 =20μF 𝐶 2 = ? C 2 =? 𝑄 = 2 × 1 0 − 4 𝐶 Q=2×10 −4 C 1. Initial Energy Stored: 𝑈 initial = 𝑄 2 2 𝐶 = ( 2 × 1 0 − 4 𝐶 ) 2 2 × 𝐶 1 = 1.1 × 1 0 − 3 𝐽 U initial = 2C Q 2 = 2×C 1 (2×10 −4 C) 2 =1.1×10 −3 J 2. Final Energy Stored: 𝑈 final = 1 2 𝑈 initial = 1 2 ( 1.1 × 1 0 − 3 𝐽 ) = 0.55 × 1 0 − 3 𝐽 U final = 2 1 U initial = 2 1 (1.1×10 −3 J)=0.55×10 −3 J 3. Relationship Between Charges and Capacitors: 𝑄 = 𝑄 1 + 𝑄 2 (no charge added to the system) Q=Q 1 +Q 2 (no charge added to the system) 𝑄 1 = 𝑄 2 ( 𝑄 ) and 𝑄 2 = 𝑄 − 𝑄 1 = 𝑄 ( 1 − 1 2 ) Q 1 = 2 Q (Q)andQ 2 =Q−Q 1 =Q(1− 2 1 ) 4. Voltages Across Capacitors: 𝑉 1 = 𝑉 2 ⟹ 𝑄 1 𝐶 1 = 𝑄 2 𝐶 2 V 1 =V 2 ⟹ C 1 Q 1 = C 2 Q 2 𝑄 2 = ( 1 − 1 2 ) 𝑄 Q 2 =(1− 2 1 )Q 𝐶 2 = 𝑄 2 𝑄 1 ⋅ 𝐶 1 C 2 = Q 1 Q 2 ⋅C 1 𝐶 2 = ( 1 − 1 2 ) ⋅ 𝐶 1 = ( 1 − 1 2 ) ⋅ 20 𝜇 𝐹 ≈ 8.3 𝜇 𝐹 C 2 =(1− 2 1 )⋅C 1 =(1− 2 1 )⋅20μF≈8.3μF Q.3: Electrostatics Problem Given: Charges: 6 𝜇 𝐶 6μC, − 2 𝜇 𝐶 −2μC, 6 𝜇 𝐶 6μC Distances: 4 cm 4cm and 8 m 8m 1. Initial Potential Energy: 𝑈 initial = 2 ( 𝑘 ⋅ 6 𝜇 𝐶 ⋅ − 2 𝜇 𝐶 4 cm ) = 2 ( 𝑘 ⋅ − 12 × 1 0 − 12 𝐶 2 0.04 m ) = − 5.4 J U initial =2(k⋅ 4cm 6μC⋅−2μC )=2(k⋅ 0.04m −12×10 −12 C 2 )=−5.4J 2. Final Potential Energy: 𝐸 final = 2.71 J − 2.7 J E final =2.71J−2.7J 3. Work Done and Distance Calculation: 𝑈 final = 𝐾 𝐸 + 𝑃 𝐸 = 0 + 𝑃 𝐸 = 2 ( 𝑘 ⋅ 6 𝜇 𝐶 ⋅ − 2 𝜇 𝐶 𝑟 ) U final =KE+PE=0+PE=2(k⋅ r 6μC⋅−2μC ) 2 ( 𝑘 ⋅ − 12 × 1 0 − 12 𝐶 2 𝑟 ) = − 2.7 J 2(k⋅ r −12×10 −12 C 2 )=−2.7J 𝑟 = 2 ( 6 𝜇 𝐶 ⋅ − 2 𝜇 𝐶 𝑟 ) = 0.08 m r=2( r 6μC⋅−2μC )=0.08m 4. Final Calculation of 𝑥 x: 𝑥 = 8 2 − 4 2 = 64 − 16 = 6.9 cm x= 8 2 −4 2 = 64−16 =6.9cm Q.4: Electric Fields and Forces Part (a): Electric Fields at Point 𝑃 P Calculate the Electric Fields: For the charge 6 𝜇 𝐶 6μC at ( − 3 , 0 ) (−3,0): ∣ 𝐸 1 ∣ = 𝑘 ⋅ 6 × 1 0 − 6 𝐶 ( 3 × 1 0 − 2 𝑚 ) 2 = 6 × 1 0 7 𝑁 𝐶 ∣E 1 ∣=k⋅ (3×10 −2 m) 2 6×10 −6 C =6×10 7 C N For the charge 4 𝜇 𝐶 4μC at ( 5 , 0 ) (5,0): ∣ 𝐸 2 ∣ = 𝑘 ⋅ 4 × 1 0 − 6 𝐶 ( 5 × 1 0 − 2 𝑚 ) 2 = 1.44 × 1 0 7 𝑁 𝐶 ∣E 2 ∣=k⋅ (5×10 −2 m) 2 4×10 −6 C =1.44×10 7 C N Resolve into Components: 𝐸 1 𝑦 = 6 × 1 0 7 𝑁 𝐶 E 1y =6×10 7 C N 𝐸 1 𝑥 = 0 E 1x =0 𝐸 2 𝑦 = 1.44 × 1 0 7 𝑁 𝐶 ⋅ ( 3 5 ) = 8.64 × 1 0 6 𝑁 𝐶 E 2y =1.44×10 7 C N ⋅( 5 3 )=8.64×10 6 C N 𝐸 2 𝑥 = 1.44 × 1 0 7 𝑁 𝐶 ⋅ ( 4 5 ) = 1.15 × 1 0 7 𝑁 𝐶 E 2x =1.44×10 7 C N ⋅( 5 4 )=1.15×10 7 C N Total Electric Field Components: 𝐸 𝑥 = 𝐸 1 𝑥 + 𝐸 2 𝑥 = 0 + 1.15 × 1 0 7 = 1.15 × 1 0 7 𝑁 𝐶 E x =E 1x +E 2x =0+1.15×10 7 =1.15×10 7 C N 𝐸 𝑦 = 𝐸 1 𝑦 + 𝐸 2 𝑦 = 6 × 1 0 7 + 8.64 × 1 0 6 = 6.864 × 1 0 7 𝑁 𝐶 E y =E 1y +E 2y =6×10 7 +8.64×10 6 =6.864×10 7 C N Part (b): Force on − 2 𝜇 𝐶 −2μC Charge at Point 𝑃 P Calculate the Force Components: 𝐹 𝑥 = − 2 × 1 0 − 6 𝐶 ⋅ 1.15 × 1 0 7 𝑁 𝐶 = − 23 × 1 0 − 6 𝑁 F x =−2×10 −6 C⋅1.15×10 7 C N =−23×10 −6 N 𝐹 𝑦 = − 2 × 1 0 − 6 𝐶 ⋅ 6.864 × 1 0 7 𝑁 𝐶 = − 137.28 × 1 0 − 6 𝑁 F y =−2×10 −6 C⋅6.864×10 7 C N =−137.28×10 −6 N Resultant Force: 𝐹 ⃗ = 𝐹 𝑥 𝑖 ^ + 𝐹 𝑦 𝑗 ^ F =F x i ^ +F y j ^ 𝐹 ⃗ = − 23 × 1 0 − 6 𝑖 ^ − 137.28 × 1 0 − 6 𝑗 ^ F =−23×10 −6 i ^ −137.28×10 −6 j ^ Magnitude: ∣ 𝐹 ⃗ ∣ = ( 𝐹 𝑥 ) 2 + ( 𝐹 𝑦 ) 2 ∣ F ∣= (F x ) 2 +(F y ) 2 Direction: 𝜃 = tan ⁡ − 1 ( 𝐹 𝑦 𝐹 𝑥 ) θ=tan −1 ( F x F y ) Part (c): Calculating Field 𝐸 𝑑 E d Given Configuration: Charges: 6 𝜇 𝐶 6μC at 𝐴 A, − 4 𝜇 𝐶 −4μC at 𝐵 B Distances: 4 cm 4cm Calculate 𝐸 𝑑 E d : For − 4 𝜇 𝐶 −4μC: 𝐸 𝑑 = 𝑘 ⋅ − 4 𝜇 𝐶 ( 0.04 𝑚 ) 2 E d =k⋅ (0.04m) 2 −4μC Net Electric Field at 𝑑 d: 𝐸 𝑑 = 𝑘 ⋅ 6 × 1 0 − 6 𝐶 ⋅ 4 ( 6 × 1 0 − 2 ) 2 ⋅ cos ⁡ ( 𝜃 ) = 1.15 × 1 0 7 𝑁 𝐶 E d =k⋅ (6×10 −2 ) 2 6×10 −6 C⋅4 ⋅cos(θ)=1.15×10 7 C N