What is the magnitude of the magnetic force on a charged particle (Q = 5.0 µC) moving with a speed of 80 km/s in the positive x direction at a point where Bx = 5.0 T, By = - %3D ?4.0 T, and Bz = 3.0 T
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- Current I = 30 A flows in a straight wire. An ion with charge Q = 3.2e-19 C is moving toward the wire, at distant r = 0.10 m, with speed v = 5.00×106 m/s. The magnetic force on the ion is antiparallel to the current. What is the magnitude of the force? (in N) OA: 3.742×10¯ OB: 4.379x10- OC: 5.123x10- OD: 5.994x10- 17 17 17 17 Submit Answer Tries 0/12 OE: 7.013x10 17 OF: 8.205x10- 17 OG: 9.600x10" OH: 1.123×10 17 16 Send Feedback 05008HISSOA charged particle with a charge to mass ratio of 5.7E8 C/kg travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.84 T. How much time does it take for the particle to complete one revolution?A laboratory electromagnet produces a magnetic field of magnitude 1.56 T. A proton moves through this field with a speed of 6.54 ✕ 106 m/s. (a) Find the magnitude of the maximum magnetic force that could be exerted on the proton. N(b) What is the magnitude of the maximum acceleration of the proton? m/s2(c) Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the same direction as the proton.) YesNo Explain. This answer has not been graded yet. (d) Would the electron undergo the same acceleration? YesNo Explain.
- This time I1 = 9.01, I2 = 1.40 A, and the two wires are separated by 8.41 cm. Now consider the charge q = 5.09 x 10^-6 C, located a distance of 7.00 cm to the right of wire I2, moving to the right at speed v = 66.0 m/s. What is the magnitude of the total magnetic force on this charge? 2.21E-09 N 1.11E-09 N 5.27E-09 N 3.69E-09 NThis time I1 = 5.03, I2 = 1.44 A, and the two wires are separated by 9.70 cm. Now consider the charge q = 7.27 x 10^-6 C, located a distance of 4.27 cm to the right of wire I2, moving to the right at speed v = 24.6 m/s. What is the magnitude of the total magnetic force on this charge? 1.19E-09 N 9.52E-10 N 2.49E-09 N 1.99E-09 NQ. 3: A proton moves through a uniform magnetic field given by B = (30 î – 205) mT. At a time t1, the proton has a velocity given by = the proton is FB (Vz î + (2000m/s)ĵ) and the magnetic force of (4 * 10-17N) k. At that instant, what is the velocity vx?
- Current I = 30 A flows in a straight wire. An ion with charge Q = 3.2e-19 C is moving toward the wire, at distant r = 0.10 m, with speed v = The magnetic force on the ion is antiparallel to the current. What is the magnitude of the force? (in N) 6.00×106 m/s. A: 3.682x10-17 B: 4.897x10-17 OC: 6.513x10-17 OD: 8.662x10-17 E: 1.152x10-16 OF: 1.532x10-16 OG: 2.038×10-16 OH: 2.710×10-16A proton (with mass mp = 1.67 x 10-27 kg, charge qp = 1.6 x 10-19 C) moving with a speed of 0.80 x 105 m/s, enters the Earth’s magnetic field (B= 55.0 μμT, Northbound). If the proton is shot eastward, the magnetic force acting on it should be Group of answer choices 8.08 x 10-16 N downward 7.04 x 10-19 N West-bound 7.04 x 10-19 N upward 7.04 x 10-19 N downwardAn electron in a magnetic field moves along a circle with a radius of 25.4 m with a speed that follows: v(t)=V0e−bt where b = 0.56 s-1and V0 = 170 m/s. I need to find the angular acceleration at t=3.7s, but I don't know how I'm supposed to get that becuase the formula that we have says that it is the second derivative of angular position, but when I calculate angular position all I get is 6.57516 radians. I don't understand how I'm supposed to take a derivative of that since it doesn't have any variables. I don't know if it is needed for this, but I already calculated the centripetal acceleration as 18.0446m/s^2