Physics NY poblem set 5 solutions 1 Physics NY poblem set 5 solution Hello eveybody, this is ED. Hi ED! ED is useful fo dawing the ight hand ule when you don t know how to daw. When you have a coss poduct such as C= A, ED lies along A, looks along vecto (note how his eyes ae the symbol fo a vecto coming out of the page) and his left am points along vecto C. Pat A Question A E! + He + He + E! - He -! + - Fo pat, the electic foce points down E Which means that the magnetic foce points up And by the ight hand ule, (shown below) the magnetic field is in the page. Question A magnetic field can only exet a foce on a moving paticle, as we can see in the fomula F = qv. So if a paticle is initially at est (i.e. not moving) the magnetic field cannot
Physics NY poblem set 5 solutions 2 exet a foce, which means it cannot acceleate the paticle, which means it cannot set the paticle in motion. Now I ask you: can a magnetic field stop a paticle? (That would be a good one fo a test, don t you think?) Question C If we hang a cuent loop fom a thead, the magnetic field will exet a toque on the cuent loop ( τ = µ ). This toque will cause the loop to tun until its magnetic dipole moment is aligned with the magnetic field. Now I ask you: what will be the toque at that moment, and what happens if the cuent loop is vey light? Question D qv= p p 2q p qv α α q = theefoe 2 α vp = v Ans: C) p Question E y E + z x F points down (see ED s am) so the electic foce, E field must point up as well. Ans: A) F, must point up and the electic
Physics NY poblem set 5 solutions 3 Question F! l Accoding to the ight hand ule as shown above, the magnetic field must point into the page Question G v! negative chage moves down qv! equivalent to positive chage moving up F The foce points Noth Question H Ans : A
Physics NY poblem set 5 solutions 4 Question I a) The Eath s magnetic field exets a foce on incoming chaged paticles, deflecting them pependicula incoming chaged paticles, deflecting them pependicula to thei velocity. As a esult, the paticle becomes tapped as it spials aound a magnetic field line. (Paticles tapped in the Eath magnetic field fom the van Allen adiation belts ) b) The Eath s magnetic field cuve downwad and become most dense as they appoach the magnetic poles. It is hee that the spialling chaged paticles ae bought closest to the planet; unde cetain conditions they come low enough to collide with atoms and molecules in the uppe atmosphee, poducing light. Question J Yes, if the chage moves paallel to the magnetic field lines thee will be no foce. F = qvbsinθ Questions K No F = qv the foce is always pependicula to the paticle s motion. Question L Let s do an ode of magnitude estimate I 5 A -4 T So the total foce on the m pole (assume to F Il 5 4 ( A)( m)( N/Am ) ) E F Il 0 N 0 N is the weight of 0 apples, that foce is unlikely to bend a flagpole.
Physics NY poblem set 5 solutions 5 Pat Question 1 a)! 5 m v = 5 s 30 o v = vcos30i ˆ+ vsin 30j ˆ v = 4.33 ˆi + 2.50 ˆj ( ) 5 5 m s F = 1.6 2.50 1.00 i + 1.00 4.33 j+ 4.33 2.00 6.00 2.50 k F ( ) 19 5 5 1.6 4.33 2.50 0 ( ˆ ˆ ˆ) ( 19 ) ( 5 ) ( 5 ) ( 5 ( ) 5 ) = 4.00i+6.93j+38.9k N is the esultant magnetic foce ( ˆ ˆ ˆ 14 ) ˆ ˆ ˆ i j k F = qv = 6.00-2.00 1.00 F = ( )( ) 19 5ˆ 5ˆ 1.6 2.50 i 4.33 j 24.3 5 k ˆ b) F = qe We want the electic foce to be 14 F ( 4.00i-6.93j-38.9k ˆ ˆ ˆ E = + ) N so that FE + F = 0 14 ( 4.00i ˆ 6.43j ˆ 38.4k ˆ) F E = = 19 q 1.60 5 N E = ( 2.50i ˆ+ 4.33j ˆ+ 24.3kˆ) is the electic field we need to make the net foce on C the electon zeo.
Physics NY poblem set 5 solutions 6 Question 2 2 N E = 8.00 C = 0.800T m = 1.16-26 kg paticle is singly chaged theefoe: q = ±1.6-19 C (Can you explain why that is so? Do you think we might ask that on a test?) a) Fist we find the velocity of the paticle in the velocity selecto: 2 E 8 3 m v = = = 1.00 0.8 s This is the final velocity of the acceleating stage. The paticle stats fom est so v i = 0 m/s U + K = U + K i i f f U U = K K i f f i Δ U = Δ K 1 2 1 2 qδ V = mvf mvi 2 2 1 2 1.16 26 3 ( 1 ) 0 Δ V = 2 = 36.2 19 1.6 ( ) We need to acceleate the paticle though a potential diffeence of 36.2mV b) 26 3 mv 1.16 1 5 = = = 9.06 m 19 q 1.6 0.8 The adius of the cicula path followed by the paticle once it is inside the chambe is: 9.06-5 m c) - E! + - E E! 3 V - + Right hand ule fo magnetic foce:
Physics NY poblem set 5 solutions 7 Question 3 a) The magnetic foce is to the ight (see pictue) I + - I Right hand ule fo magnetic foce: b) Given : I=0.5kg l =0.1 m =5.00T v o =0 m/s t = 3 s Objective : a Stategy : calculate F, find acceleation and use kinematics to find the final speed F = Il sinθ whee θ = 90 F = 4 0.1 5 = 2.00N F = ma F 2N m a = = = 4.00 2 m 0.5kg s m v= vo + at = 0+ 4 3= 12.0 s The ba will have a speed of 12 m/s 3 seconds afte stating fom est
Physics NY poblem set 5 solutions 8 Question 4 Appopiate ight hand ules ae descibed by ED s I I Fee body diagam T! G List of foces on the y axis : T =? F = mg G F = Ilsinθ y Newton s second law, the sum of the foces should be zeo T F F = 0 G T = FG + F = mg+ Il sinθ = 0.3 9.8 + 3 0.1 2.5 sin 90 T = 3.69N The tension in the sting suppoting the loop is 3.69 N in positive j diection.
Physics NY poblem set 5 solutions 9 Question 5 This poblem will be easie if we look at the diagam fom the positive y-axis: I out x y z I in a) F ab = Ilsinθ = 5 0.05 0.5 sin90 F ab = 0.125 N towads the left which we can also wite as : I in F = 0.125i ˆ N ab F cd = Ilsinθ = 5 0.05 0.5 sin90 F cd = 0.125 N towads the ight which we can also wite as: F = 0.125i ˆ N cd I out F da = Ilsinθ = 5 0.05 0.5 sin30 F da = -0. 625 N into the page which we can also wite as: F = 0.0625j ˆ N da
Physics NY poblem set 5 solutions I da 30 o F bc = Ilsinθ = 5 0.05 0.5 sin150 F bc = 0.0625 N out of the page which we can also wite as: F = 0.0625j ˆ N bc I bc 150 o b) µ = NIA = 1 5 0.05 = 0.0125Am τ = µ sinθ 2 2 τ = = τ = 5.41 3ˆ j Nm 3 0.125 0.5 sin120 5.41 Nm I out x y µ! z 120 o τ! I in The value of the toque is 5.41-3 Nm pointing in the negative y-diection.
Physics NY poblem set 5 solutions 11 Question 6 Given : «doubly chaged positive ion means» q = 2 1.6-19 C d = 0.2000 m E = 5.00 4 V/m = 0.5 T = 0.500 cm E! ++ + - mv In the magnetic field, the adius of the paticle s tajectoy is : = q We know, q, so we need to find v to get m Δ K = Δ U Δ K = qδ V 1 2 1 2 mv f mvi = qed 2 2 1 2 mv f = qed 2 2 2qEd v = m 2 2 2 mv = 2 2 q 2 2 m 2qEd = 2 2 qm 2 ( ) ( ) 2 2 2 3 19 2 q 5 2 1.6 0.5 m= = = 1.00 4 2qEd 2 5 0.2 So the mass of the paticle is 1.00-28 kg 28 kg
Physics NY poblem set 5 solutions 12 Question 7 I c I R coil R Stategy: R and the coil ae in paallel. If we find I C, the cuent in the coil, we will know the voltage acoss the coil V C and theefoe acoss the combination. Fom V and I we will get R EQ. 6 N m 4 τ = 5 40deg= 2.00 Nm deg τ = NI A sin90 I C V V C 4 τ 2 = = = 2 2 NA 500 0.01 7.5 2 = I R = 5.33 14 = 0.746 V C C COIL C = IR EQ V 0.746 REQ = = = 1.49 Ω I 0.500 The total esistance of the ammete 1.49 Ω 2 5.33 A