A spark can jump between two nontouching conductors if the potential difference between them is sufficiently large. A potential difference of approximately 940 V is required to produce a spark in an air gap of 1.0 x 104 m. Suppose the light bulb in the figure is replaced by such a gap. How fast would a 2.55-m rod have to be moving in a magnetic field of 6.03 T to cause a spark to jump across the gap? Number Units Conducting rail be
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- A wire has a current of 45 A going into the page. What is the B field and direction 1.5 m to the right of the wire? What is the force and direction on a proton traveling 5 m/s into the page at this point?A particle passes through a mass spectrometer as illustrated in Figure P19.36. The electric field between the plates of the velocity selector has a magnitude of 8 250 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.093 1 T. In the deflection chamber the particle strikes a photographic plate 39.6 cm removed from its exit point after traveling in a semicircle. (a) What is the mass-to-charge ratio of the particle? (b) What is the mass of the particle if it is doubly ionized? (c) What is its identity, assuming it’s an element?A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field between the plates. The magnetic field is parallel to the plates and perpendicular to the initial direction of the protons, and its magnitude is 2.3 T. The proton beam passes through undeflected. What is the speed of the protons?_______ m/s
- A velocity selector in a mass spectrometer uses a 0.130 T magnetic field. a. What electric field strength (in volts per meter) is needed to select a speed of 3.70 ✕ 106 m/s? b. What is the voltage (in kilovolts) between the plates if they are separated by 1.00 cm?An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.230 T. If the kinetic energy of the electron is 3.70 x 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron 9.02e5 ✔ m/s (b) the radius of the circular path Your response is off by a multiple of ten. umTwo wires are located at the distance d=4m as shown in the diagram. y d/2 #1 #2 d/2 A d/2 d/2 B Wire #1 carries a current of 4 A in the shown direction. Wire #2 carries a current of 2 A in the shown direction. Hol Units: Tesla (T) B = 2nr' Ho=47tx10-7 T*m/A 1. Find the magnitude and direction of total magnetic field at point A, located exactly in the middle between two wires. 2. Find the magnitude and direction of total magnetic field at point B, located at the distance d/2 to the right of wire #2 as shown in diagram. 3. Find the magnitude of total magnetic field at point C, located at the distance d/2 above the point A as shown in diagram. 4. Find the direction (angle related to positive +x axis) of total magnetic field at point C, located at the distance d/2 above the point A as shown in diagram.
- Consider a wire carrying a current due east in a location where the Earth’s field is due north. A) What is the direction of the force on the wire if both are parallel to the ground? B) Calculate the force per unit length on the wire, in newtons per meter, if the wire carries 19.5 A and the field strength is 2.85 × 10-5 T. C) What would the diameter be, in meters, of copper wire (with density ρ = 8.80 × 103 kg/m3) that would have its weight supported by this force in meters? D) Calculate the resistance per unit length, in ohms per meter, that this copper wire will have. Copper has a resistivity of ρR = 1.72 × 10-8 Ω⋅m. E) Calculate the potential difference per unit length, in volts per meter, of this wire.See attached imageA square, 44.0-turn coil that is 13.0 cm on a side with a resistance of 0.900 2 is placed between the poles of a large electromagnet. The electromagnet produces a constant, uniform magnetic field of 0.400 T directed into the screen. As suggested by the figure, the field drops sharply to zero at the edges of the magnet. The coil moves to the right at a constant velocity of 3.10 cm/s. N-turn coil B into the screen V