homework-11-sln

Solution for Wednesday Homework 11 Magnetism (Required but not Collected) Solution to Wednesday Homework Problem 11.1(Basic Biot-Savart) Problem: To the right is the end view of a wire carrying current out of the page. Consider a small chunk of the current, Id vector ℓ . Use the Biot-Savart law to calculate the direction of the magnetic field, d vector B A , at point A . A Current out of Page I y x Select One of the Following: (a) The field at A due to the small element of current points to the top of the page. (b) The field at A due to the small element of current points to the bottom of the page. (c-Answer) The field at A due to the small element of current points to the left of the page. (d) The field at A due to the small element of current points to the right of the page. (e) The field at A due to the small element of current points out of the page. (f) The field at A due to the small element of current points into the page. Solution A B C B dB A dB B dB C r A r B r C I Definitions Id vector ℓ ≡ Current Element out of Page vector r i ≡ Vector from Wire to Point i d vector B i ≡ Magnetic Field at Point i The Biot-Savart Law states that at point A, the magnetic field from current element Id vector ℓ is given by d vector B A = parenleftbigg μ 0 4 π parenrightbigg Id vector ℓ × ˆ r A r 2 A . The direction of the field is given by the cross-product of the direction of the current with the direction of the displacement of the current to the field point. The direction of Id vector ℓ is out of the page and the direction of vector r A 1

points from the wire to point A is to the top of the page. Apply the Right Hand Rule to get the direction of d vector B A . Point the fingers of the right hand in the direction of Id vector ℓ (out of the page) and curl the fingers in the direction of vector r A ; this leaves the pointing to the left, or the direction drawn for d vector B A . Total Points for Problem: 3 Points SolutiontoWednesdayHomeworkProblem11.2(Ampere’sLawwithThickConductor) Problem: The system of conductors shown to the right is cylindrically symmetric. It is composed of three conductors A , B , and C . The center con- ductor, A , is a thin wire which carries a current 3 I out of the page. The second conductor is a thick conducting tube with inner radius a and outer ra- dius b . It carries a current I into the page which is distributed uniformly in its interior. The third conductor C is a thin tube which carries a current 2 I into the page. Compute the magnitude of the magnetic field in Region I. Select One of the Following: (a) B I = 0 (b-Answer) B I = 3 μ 0 I 2 πr (c) B I = 3 μ 0 Ir 2 πa 2 (d) B I = 2 μ 0 I ( r 2 - a 2 ) 2 π ( b 2 - a 2 ) (e) B I = 2 μ 0 I 2 π ( b - a ) A B C I II III IV Solution The current enclosed by an Amperian path placed in region I is the current of conductor A , which is 3 I . Ampere’s Law gives the field B in region I is B I = μ 0 I enc 2 πr = 3 μ 0 I 2 πr Total Points for Problem: 10 Points Solution to Wednesday Homework Problem 11.3(Direction of Current Induced in a Solenoid by a Changing Resistance in Another Solenoid) Problem: The figure to the right shows two solenoids. Solenoid 1 is powered by a battery connected to a resistor with resistance R . The two ends of Solenoid 2 are connected through a second resistor. If the resistance of the resistor connected to Solenoid 1 is increased over some period of time, does current flow in Solenoid 2 ? If so, in what direction? a b 1 2 + _ 2

Total Points for Problem: 10 Points Solution to Wednesday Homework Problem 11.5(Long Time Behavior of RL Circuit) Problem: A battery with potential difference Δ V is connected in series to an inductor with inductance L and a resistor with resistance R . What is the current flowing in the circuit a long time after the battery is connected? Select One of the Following: (a) 0 (b) Δ V/L (c-Answer) Δ V/R (d) L/R (e) LR Solution After a long time, the current in an RL circuit is no longer changing and therefore the potential difference across the inductor is zero. The loop equation for the circuit becomes Δ V - IR - 0 = 0 , and therefore the current in the circuit is I = Δ V R . Total Points for Problem: 2 Points Solution to Wednesday Homework Problem 11.6(Multiple Approximation with Earth’s Magnetic Field Solenoid) Problem: Let’s work with the solenoid used to measure the earth’s magnetic field. The solenoid was wound with 79 turns over a distance of 79cm . The diameter was d = 11cm . A TA mistakenly runs the wire feeding the solenoid back across the end of the solenoid and along the top. This problem asks you to analyze the error created in the experiment by this foolish setup. The experiment measures the magnetic field at point P at the center of the solenoid. The distance from the wire segment to the point P is ℓ = 0 . 5m . Since the percent error will be the same for any current, let’s analyze the system at the maximum output of out power supplies I = 17A . Note the coordinate system for the problem has the plane of the paper as the x - z plane. (a)On the figure below draw the direction of the magnetic field of the solenoid, vector B sole , the wire segment from A to B, vector B AB , and the infinite straight wire from B to C, vector B BC , at point P . (b)Calculate the field of the solenoid at point P in the infinite solenoid approximation. Report the field as a vector. (c)Calculate the field of the wire segment AB in the finite current element approximation at point P . Report the field as a vector. (d)Calculate the field of the wire segment BC in the approximation that it is an infinite straight wire at point P . Report the field as a vector. (e)Calculate the magnetic field of the segment BC exactly at point P . Report the field as a vector. Report both a symbolic and numeric answer. Use the point P as the origin for this calculation. (f)Does the TA bungle introduce more than 10% relative error in the experiment? 4

A B C d P I x z Solution to Part (a) The direction of the field can be found using the right hand rule for a wire in each case. A B C d P I x z y B AB B BC B sole Grading Key: Part (a) 3 Points Solution to Part (b) The magnetic field of an infinite solenoid is B = N L μ 0 I = 79 0 . 79m (4 π × 10 - 7 Tm A )(17A) = 0 . 00214T Still pathetic. The direction of the field is - ˆ x . vector B sole = - 0 . 00214Tˆx 5

(a)Set up the integral: Divide the wire into small segments of length Δ x . The location of P is vector r P = (0 , 0 , 0) . The location of the i th segment is vector r i = ( x i , 0 , d/ 2) . The displacement vector from the i th segment to the point P is vector r iP = vector r P - vector r i = ( - x i , 0 , - d/ 2) . The length of the displacement vector is r iP = radicalBig x 2 i + ( d 2 ) 2 . The total magnetic field is found by summing the fields of the segments vector B P = summationdisplay i μ 0 I 4 π Δ vector ℓ × vector r iP r 3 iP The vector Δ vector ℓ points in the direction of current, Δ vector ℓ = - Δ x ˆ x . Evaluate the required cross-product Δ vector ℓ × vector r iP = ( - Δ x ˆ x ) × ( - x i , 0 , - d/ 2) = Δ xd 2 ˆ x × ˆ z = - d 2 ˆ y Δ x The total field at P is then vector B P = μ 0 I 4 π summationdisplay i Δ vector ℓ × vector r iP r 3 iP = μ 0 I 4 π summationdisplay i - d 2 ˆ y Δ x ( x 2 i + ( d 2 ) 3 2 = - μ 0 I 4 π d 2 ˆ y summationdisplay i Δ x ( x 2 i + ( d 2 ) 3 2 Convert to an integral vector B P = - μ 0 I 4 π d 2 ˆ y integraldisplay ℓ - ℓ dx ( x 2 + ( d 2 ) 2 ) 3 2 x z P d/2 i I - r iP (b) Do the integral: The integral is even vector B P = - 2 μ 0 I 4 π d 2 ˆ y integraldisplay ℓ 0 dx ( x 2 + ( d 2 ) 2 ) 3 2 Perform the integral integraldisplay ℓ 0 dx ( x 2 + ( d 2 ) 2 ) 3 2 = x ( d 2 ) 2 radicalBig x 2 + ( d 2 ) 2 vextendsingle vextendsingle vextendsingle vextendsingle ℓ 0 Substitute the limits integraldisplay ℓ 0 dx ( x 2 + ( d 2 ) 2 ) 3 2 = ℓ ( d 2 ) 2 radicalBig ℓ 2 + ( d 2 ) 2 Put it back together vector B P = - 2 μ 0 I 4 π d 2 ˆ y ℓ ( d 2 ) 2 radicalBig ℓ 2 + ( d 2 ) 2 = - μ 0 I dπ ˆ y ℓ radicalBig ℓ 2 + ( d 2 ) 2 which does the right thing as ℓ → ∞ so we’re cool. Now, calculate the field vector B P = - μ 0 I dπ ˆ y ℓ radicalBig ℓ 2 + ( d 2 ) 2 = - (4 π × 10 - 7Tm A )(17A) (0 . 11m) π ˆ y (0 . 395m) radicalBig (0 . 395m) 2 + ( 0 . 11m 2 ) 2 = - 6 . 12 × 10 - 5 Tˆy Grading Key: Solution to Part (e) 5 Points Solution to Part (f) 7

The total field of the wire from A to C is not near 10% of the solenoid field, so the miswiring is not significant. Grading Key: Part (e) 1 Points Total Points for Problem: 17 Points Solution to Wednesday Homework Problem 11.7(Ampere’s Law with Copper Pipe) Problem: A copper water pipe has an inner radius a = 1 . 384cm and an radius diameter b = 1 . 588cm . If the pipe is long compared to its radius, it can be modeled as an infinite cylindrical conductor. Let a current I flow down the pipe such that the current is uniformly distributed through the copper. The current flows into the page. An end view of the system is drawn below. (a)Draw the magnetic field in all regions. (b)Calculate the magnetic field symbolically in all regions. (c)Suppose the pipe carries a current I = 10A . What is the magnetic field at a point P at 1 . 486cmˆx as drawn below? Let the origin be along the axis of the pipe. Report the field as a vector. I II III a b P Air Air y x copper Solution to Part (a) A total current into the page is encircled in regions II and III producing a clockwise field, by the right-hand rule for the wire. The total current encircled in region I is zero, so the field in region I is zero. I II III Air Air y x 8

Problem: The figure to the right shows a square coil of wire with N turns and edge length w in a magnet that produces a uni- form field as drawn. The magnetic field is constant and has magnitude B . At t = 0 , the right edge of the coil is at the right edge of the magnet and therefore approximately at the edge of the field. The coil is jerked out the the field so that the length of the coil in the field is given by ℓ ( t ) = w - γt 3 . Be careful, w is the width of the coil not the angular frequency ω . Because the loop is not rotating, the angular frequency is ω = 0 . (a)Calculate the magnetic flux through the loop as a function of time sym- bolically. (b)Calculate the emf as a function of time symbolically. (c)Draw the direction of the induced current. Justify your choice of di- rections physically. w Uniform Magnetic Field Out of Page v (t) Solution to Part (a) φ m = NBA = NBwℓ ( t ) = NBw ( w - γt 3 ) Grading Key: Part (a) 10 Points Solution to Part (b) The emf is emf = - dφ w dt = 3 γNBwt 2 by Faraday’s law. Grading Key: Part (b) 10 Points Solution to Part (c) Flux is decreasing out of the page, so the induced current acts to create a field in the same direction as the existing field. The current, then, is counterclockwise. Grading Key: Part (c) 10 Points Total Points for Problem: 30 Points Solution to Wednesday Homework Problem 11.9(Generator Problem) Problem: A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, vector B = B 0 ˆ z = 0 . 1Tˆ z , such that the normal of the coil is aligned with the field once per rotation. The coil rotates once every T = 5s and starts with its normal parallel to the field. 10

(a)Write the magnetic flux as a function of time symbolically. Report the numeric values for any constants you introduce separately. (b)Compute the induced emf at t = 12 . 5s . Report both a symbolic and numeric value. Solution to Part(a) The flux is defined as φ m = NA vector B · ˆ n , where N = 30 is the number of turns and A = (2cm) 2 = 4 × 10 - 4 m 2 is the area of the coil. vector B · ˆ n = B cos( ωt + δ ) , where ω is the angular frequency and δ = 0 , since the normal is parallel to the field at t = 0 . The loop rotates once in 5s , so its frequency is f = 1 / 5s = 1 5 Hz . The angular frequency is then ω = 2 πf = 2 π 5 Hz . The flux is therefore, φ m = NAB cos( ωt ) , with N = 30 , A = 4 × 10 - 4 m 2 , B = 0 . 1T , ω = 2 π 5 s - 1 . Grading Key: Part (a) 20 Points Solution to Part(b) The emf is found by applying Faraday’s Law to the magnetic flux computed above: emf ( t ) = - dφ m dt = - NAB d cos( ωt ) dt = NABω sin( ωt ) so emf (12 . 5 s ) = (30)(4 × 10 - 4 m 2 )(0 . 1T)( 2 π 5 Hz) sin(( 2 π 5 Hz)(12 . 5s)) = 0V . Grading Key: Part (b) 10 Points Total Points for Problem: 30 Points 11