For problem 5, calculate the x component ofthe magnetic field in Tesla using a velocity in m/sof 2.0 i + 4.5j. (5 sig. figs) 5 GO An electron moves through a uniform magnetic field givenby B = B¸î + (3.0B.)ĵ. At a particular instant, the electron has ve-locity v(2.01 + 4.0)) m/s and the magnetic force acting on it is(6.4 × 10-19 N)k. Find B¸.= QUESTION 1For problem 28.5, calculate the x component ofthe magnetic field in Tesla using a velocity in m/sof 2.0 i +4.5 j. 5 sig. figs.

Answered: Image /qna-images/answer/b157172d-af47-4ff4-bf32-4ec9b447f689.jpg

Answered: 5 GO An electron moves through a…

Similar questions

A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a speed of 2.96 ✕ 105 m/s and enter a uniform magnetic field of 0.605 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by singly charged ions of 235U and 238U??? cm
A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various speeds. If the electric field strength is 8.4 x 10oª N/C, what should be the value of the magnetic field (in tesla) to select protons of velocity 4.2 x 105 m/s? Mass of proton = 1.67 x 10-27 kg. Charge of proton = 1.6 x 10-19 C.
A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from a velocity selector at a speed of 3.00 x 105 m/s and enter a uniform magnetic field of 0.375 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by ions of 235U and 238U? The mass of a 235U ion is 3.902 x 10-25 kg, and the mass of a 238U ion is 3.952 x 10-25 kg, while both types have a charge of 1.602 x 10-19 C. A mass spectrometer (schematic). Charged particles are first sent through a velocity selector. They then enter a region where a magnetic field B causes positive ions to move in a semicircular path and strike a photographic film at P.
A proton (H+) moves at 5.00E+07 m/s perpendicular to a magnetic field. The ion enters perpendicular to a large magnetic field of strength of 2.50 T so that the proton cannot leave the magnetic field. (a) What is the radius of the circular motion of the proton, in meters? (b) What is the period of the circular motion, in nanoseconds? (c) What is the frequency of the motion, in MHz?
A short current element di = (0.500 mm) carries a current of 5.30 A in the same direction as di Point P is located at 7 = (-0.730 m)i + (0.390 m)k. Find the magnetic field at P produced by this current element. Enter the x, y, and z components of the magnetic field in teslas separated by commas. dBx, dBy, dB₂ = VE ΑΣΦ ? T
An ion carrying a single positive elementary charge (+e) has a mass of 4.22x10-27 kg. It is accelerated through an electric potential difference of 1.29 kV and then enters a uniform magnetic field of B = 0.2 T along a direction perpendicular to the field. What is the radius of the circular path of the ion in the magnetic field? (Units: cm)
A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a speed of 2.78 x 105 m/s and enter a uniform magnetic field of 0.575 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by singly charged ions of 235U and 238U? cm Need Help? Read It
ASAP
A uniform electric field of 7.42 V/m west and a uniform magnetic field of 7.78 T west exist in a region of space. Find the magnitude of the acceleration of an electron moving with velocity 1.52 m/s north in this region. Answer in units of m/s2.
An electron moving at the speed of 107 m/s, enters a magnetic field of induction 0.1 T at right angle to the velocity of electron. The radius of circular path followed by the electron is

5 GO An electron moves through a uniform magnetic field given by B = B¸î + (3.0B)ĵ. At a particular instant, the electron has ve- locity v (2.01 + 4.0)) m/s and the magnetic force acting on it is (6.4 x 10-19 N)k. Find B.. =