hw5 oscar - maldonado (om2892) – Homework 4 2010 UPII –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: maldonado (om2892) – Homework 4 2010 UPII – Stone – (13750) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two identical parallel sections of metal rods are connected parallel to a battery as shown. The two sections of metal rods are free to move. b b 5 . 1m . 24 m 2 . 8 V Note: Neglect the resistance of the part of the circuit not specifically mentioned below. The battery is 2 . 8 V, each metal rod has a resistance of 2 . 04 Ω, a length of 5 . 1 m, and a mass of 0 . 0024 kg . The rods are separated by . 24 m . Find the magnitude of the acceleration of each rod. Correct answer: 0 . 00333606 m / s 2 . Explanation: Let : E = 2 . 8 V , R = 2 . 04 Ω , ℓ = 5 . 1 m , m = 0 . 0024 kg , and Δ r = 0 . 24 m . b b ℓ Δ r E The attractive forces on each metal rod are equal and opposite to each other, and each force has its magnitude given by F = I ℓ B with I the current in a rod, ℓ the length of that rod and B the magnetic field at the rod due to the current in the other rod. Since the separation Δ r of the rods, is much less than ℓ we can use Ampere’s law to find contintegraldisplay vector B · d vector ℓ = 2 π Δ r B = μ I B = μ I 2 π Δ r . Further, the voltage drop through the rod equals the voltage E of the battery; i.e. , E = I R , and I = E R , and B = μ parenleftbigg E R parenrightbigg 2 π Δ r . Substituting I = E R , and the expression for B , gives for the force, F = I ℓ B = parenleftbigg E R parenrightbigg 2 μ ℓ 2 π Δ r With F = ma , we find a = parenleftbigg E R parenrightbigg 2 μ ℓ 2 π Δ r m = parenleftbigg 2 . 8 V 2 . 04 Ω parenrightbigg 2 (1 . 25664 × 10 − 6 N / A 2 )(5 . 1 m) 2 π (0 . 24 m)(0 . 0024 kg) = . 00333606 m / s 2 . 002 (part 1 of 3) 10.0 points A circular current loop of radius R is placed in a horizontal plane and maintains a current I . There is a constant magnetic field vector B in the xy-plane, with the angle α ( α < 90 ◦ ) defined with respect to y-axis. The current in the loop flows counterclockwise as seen from above. maldonado (om2892) – Homework 4 2010 UPII – Stone – (13750) 2 I z ˆ k x ˆ ı y , ˆ B α What is the direction of the magnetic mo- ment vectorμ ? 1. +ˆ ı 2.- sin α ˆ - cos α ˆ ı 3. sin α ˆ + cos α ˆ ı 4. +ˆ correct 5.- cos α ˆ + sin α ˆ ı 6. cos α ˆ - sin α ˆ ı 7.- ˆ k 8. + ˆ k 9.- ˆ ı 10.- ˆ Explanation: I z ˆ k τ x ˆ ı μ loop B α According to the right-hand rule, the mag- netic moment is upward. 003 (part 2 of 3) 10.0 points If the current in the loop is 0 . 235 A and its radius is 1 . 67 cm, what is the magnitude of the magnetic moment of the loop?...
View Full Document

This note was uploaded on 03/31/2011 for the course PHYS 1001 taught by Professor Stone during the Summer '10 term at Texas Brownsville.

Page1 / 9

hw5 oscar - maldonado (om2892) – Homework 4 2010 UPII –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online