A current-carrying wire is oriented along the x axis. It passes through a region 0.22 m song in which there is a magnetic field of 2.44 T in the +z direction. The wire experiences a force of 13 N in the +y direction as shown in the figure. 1. 2. 3. 4. O O O O (2.44) (13)x (0.22) (13) (2.44)x (0.22) (0.22) (13) x (2.44) O The magnetic field vector acting on the wire can be expressed as: >ITI The magnitude of the electric current can be calculated using: ODownward OThe right OThe left OUpward OTO Olo ΟΙΟ ΟΤΟ B The magnitude of the current is equal to: /= The direction of the current flow in the wire will be pointing to: Downward OThe right OThe left 0 5. If the current direction is opposite to the calculated one, the magnetic force. direction acting on the wire will be pointing to: OUpward

A current-carrying wire is oriented along the x axis. It passes through a region 0.22 m song in which there is a magnetic field of 2.44 T in the +z direction. The wire experiences a force of 13 N in the +y direction as shown in the figure. 1. 2. 3. 4. O O O O (2.44) (13)x (0.22) (13) (2.44)x (0.22) (0.22) (13) x (2.44) O The magnetic field vector acting on the wire can be expressed as: >ITI The magnitude of the electric current can be calculated using: ODownward OThe right OThe left OUpward OTO Olo ΟΙΟ ΟΤΟ B The magnitude of the current is equal to: /= The direction of the current flow in the wire will be pointing to: Downward OThe right OThe left 0 5. If the current direction is opposite to the calculated one, the magnetic force. direction acting on the wire will be pointing to: OUpward

Physics for Scientists and Engineers

10th Edition

ISBN:9781337553278

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter29: Sources Of The Magnetic Field

Section: Chapter Questions

Problem 8P: One long wire carries current 30.0 A to the left along the x axis. A second long wire carries...

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One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of 2.00 C is moving with a velocity of 150iMm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undetected. Calculate the required vector electric field.
Two long, straight, parallel wires carry currents that are directed perpendicular to the page as shown in Figure P30.9. Wire 1 carries a current I1, into the page (in the negative z direction) and passes through the x axis at x = +. Wire 2 passes through the x axis at x = 2a and carries an unknown current I2. The total magnetic field at the origin due to the current-carrying wires has the magnitude 20I1(2a). The current I2 can have either of two possible values, (a) Find the value of with the smaller magnitude, stating it in terms of I1, and giving its direction. (b) Find the other possible value of I2.
A proton travels with a speed of 3.00 106 m/s at an angle of 37.0 with the direction of a magnetic field of 0.300 T in the +y direction. What are (a) the magnitude of the magnetic force on the proton and (b) its acceleration?
A wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0, (b) 90.0, and (c) 120.
A proton (charge + e, mass mp), a deuteron (charge + e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated from rest through a common potential difference V. Each of the particles enters a uniform magnetic field B, with its velocity in a direction perpendicular to B. The proton moves in a circular path of radius rp. In terms of rp, determine (a) the radius rd of the circular orbit for the deuteron and (b) the radius ra for the alpha particle.
A mass spectrometer (Fig. 30.40, page 956) operates with a uniform magnetic field of 20.0 mT and an electric field of 4.00 103 V/m in the velocity selector. What is the radius of the semicircular path of a doubly ionized alpha particle (ma = 6.64 1027 kg)?
Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46