Question
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by stepSolved in 2 steps
Knowledge Booster
Similar questions
- When a charged particle moves perpendicularly to the direction of a uniform magnetic field, the direction of the magnetic force is perpendicular to both the direction of the magnetic field and the direction of the velocity of the charged particle. Accordingly, the charge begins to move in a circular path. Since the direction of the magnetic force is toward the center of the circular path the charged particle moves along, the magnetic force is a centripetal force (recall centripetal forces from chapter 5 of the text, covered in PHY 2010). A schematic of such motion is shown below. The "X"s represent the magnetic field that is directed into the plane of this screen. The direction of velocity and magnetic force lie within the plane of this screen and are at right angles to one another, as shown below. If a charge of magnitude 4.96x10-17C, with speed 3.74x106 m/s, and mass 5.72x10-25kg moves within the magnetic field of magnitude 2.79x10-2T, what is the resulting radius of the path…arrow_forwardpart A For a particle of mass mmm and charge qqq moving in a circular path in a magnetic field BBB, find how does its kinetic energy depend on the radius of the curvature of its path r. part B Find angular momentum of a particle about the center of the circle. Express your answer in terms of the variables q, B, and r.arrow_forwardAt time t1, an electron is sent along the positive direction of an x axis, through both an electric field È and a magnetic field B ,with È directed parallel to the y axis. The figure gives the y component Fnety of the net force on the electron due to the two fields, as a function of the electron's speed v at time t1. The scale of the velocity axis is set by vs = 110 m/s. The x and z components of the net force are zero at tj. Assuming By = 0, find (a) the magnitude of the electric field and (b) B in unit-vector notation. 1 -2 v (m/s) (a) Number i Units (b) i i+ i i k Units Fnet,y (10-19 N)arrow_forward
- q5-A long straight wire is placed on the X-axis. An electric current of 49 A flows through the wire in the +X direction. At some instant of time, an electron is at the position shown in the figure, i.e. in the X-Y plane at a distance 4 cm from the wire, and is moving in the -Y direction (i.e straight towards the wire ) with a speed of V = 4.2x105 m/s. Calculate the magnitude of the force on the electron at this point due to the current in the wire and write your answer in canvas in terms of 10-18 N. On your paper also write the direction of the force. electron Long wire with currentarrow_forwardSolve the following: A proton (of mass, m-1.67 x 1027kg) starts moving in a circular path of radius 5 cm when it enters perpendicular to the Earth's magnetic field of 1.75 x 10-5 T. Calculate Bin the speed X. of the cosmic ray proton in (m/s) ? X +q tuarrow_forwardConsider a large hot air balloon that is drifting due south at a speed of 17 m/s over an area of the planet where the Earth's magnetic field of 8.9 × 10-5 T points directly up (out of the page). This provides a magnetic force of 4.8 × 10-8 N. Provide the direction of the magnetic force that is experienced by the balloon, assuming the charge buildup is (+). South West East Into the page Northarrow_forward
- An iron bolt of mass 58.5 g hangs from a string 37.5 cm long. The top end of the string is fixed. Without touching it, a magnet attracts the bolt so that it remains stationary, but is displaced horizontally 25.0 cm to the right from the previously vertical line of the string. Draw a free-body diagram of the bolt. Find the tension in the string. Find the magnetic force on the bolt. magnitude directionarrow_forwardA particle with a charge of - 5.30 nC is moving in a uniform magnetic field of B = −( 1.28 T )k. The magnetic force on the particle is measured to be F = −( 4.00×10-7 N )i +( 7.60×10−7 N )j. Part B Calculate the x component of the velocity of the particle. Express your answer in meters per second to three significant figures. ► View Available Hint(s) V₂ = -112 m/s Submit Part C Vy Previous Answers Correct Correct answer is shown. Your answer -113.09 m/s was either rounded differently or used a different number of significant figures than required for this part. Calculate the y component of the velocity of the particle. Express your answer in meters per second to three significant figures. ► View Available Hint(s) = VE ΑΣΦ = m/sarrow_forwardA magnetic field can force a charged particle to move in a circular path. Suppose that an electron moving in a circle experiences a radial acceleration of magnitude 3.0 × 1014 m/s2 in a particular magnetic field. (a) What is the speed of the electron if the radius of its circular path is 0.14 m? (b) What is the period of the motion?arrow_forward
- When a charged particle moves perpendicularly to the direction of a uniform magnetic field, the direction of the magnetic force is perpendicular to both the direction of the magnetic field and the direction of the velocity of the charged particle. Accordingly, the charge begins to move in a circular path. Since the direction of the magnetic force is toward the center of the circular path the charged particle moves along, the magnetic force is a centripetal force (recall centripetal forces from chapter 5 of the text, covered in PHY 2010). A schematic of such motion is shown below. The "X"s represent the magnetic field that is directed into the plane of this screen. The direction of velocity and magnetic force lie within the plane of this screen and are at right angles to one another, as shown below. If a charge of magnitude 7.45x1017C, with speed 4.66x106 m/s, and mass 7.12x10 25kg moves within the magnetic field of magnitude 3.96x102T, what is the resulting radius of the path (in…arrow_forwardA particle with charge q and and speed venters a region of magnetic field B1 and is deflected from its path into a region with field B2. It enters the second region moving vertically and exits horizontally. a) Draw a sketch of the path of the particle. b)The particle exits the second region at a height d, measured from the interface of the two regions. What is the mass of the particle?arrow_forwardA 100-kg metallic ball at rest is being pulled by the strange magnetic field of a planet, where the force (in newtons) is given by the km function F (r) = 4p5 Where: r = distance from the planet's surface in km m = object's mass in kg k = 5.00 km5/kg The metallic ball starts at a very distant position from the planet (r → c0) and falls towards the planet. What is the potential function U (r) of the planet? Use the condition lim,. U(r) = 0. 'r→∞ What is the metallic ball's speed (in m/s) once the steel ball is one kilometer above the planet's surface? Justify your answer using your rationale and equations used.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios