A velocity selector consists of electric and magnetic fields described by the expressions  E= Ek  and  B= B ĵ, with B = 17.0 mT. Find the value of E such that a 800-eV electron moving in the negative x direction is undeflected.  kV/m

What is the magnitude and direction of the acceleration on the electron in a 3T magnetic field ( -k) and an 800 V/m (-j) electric field, shown below (neglecting gravity), if its velocity is 300 m/s? X X X B= 3.0 T X X X X X X X X X X E=800 V/m 3.512 EE+13 m/s2 (j) -3.512 EE+13 m/s2 (j) 3.512 EE-13 m/s2 (k) 2.489 EE+13 (j) -2.489 EE+13 (j)

A pellet which holds a charge of 10 coulombs is moving upwards (+Y) and driven by an electric field in the same direction with a magnitude of 50 V/m. There is a magnetic field with a magnitude of 25 Tesla’s pointing downwards (-Y). How fast does the pellet need to be going for the magnetic force to cancel the electric force? Group of answer choices 0.5 m/s 2 m/s 4 m/s The forces will cancel out at any speed the pellet may be travelling In this case, the electric force cannot be canceled out by the magnetic force

direct answer please

Given: B = 5 * 10-5 T ẑ; σ = 4 (Ohm-meters)-1 (conductivity) a) Assume that seawater is moving at a constant velocity v = v0 ŷ and that the Earth’s magnetic field is along the ẑ-direction. Calculate the electric current density J produced by the magnetic force. Hint: first compute the force per unit charge, F/q, and then use the relationship J = σ(F/q).  b) Derive the equation of motion of a cylindrical differential volume element of base area δA and height δh parallelto the direction of J. Assume that seawater has a known volumetric mass density ρ. Show that this equation implies that the velocity satisfies the following differential equation:dvy/dt = vy/τwhere τ is a constant that you should write in terms of B, σ, and ρ.

Crossed electric and magnetic fields have magnitudes of 4.65 V/m and 74.5 mT. Find the speed at which electrons moving perpendicular to both fields can travel through them without being deflected?

A strip of copper is placed in a uniform magnetic field of magnitude 3.5 T. The Hall electric field is measured to be 2.0 x 10-3 V/m. (a) What is the drift speed of the conduction electrons m/s)? m/s (b) Assuming that n = 8.0 x 1028 electrons per cubic meter and that the cross-sectional area of the strip is 4.4 x 10-6 m², calculate the current in the strip (in A). (in m³/C)? nq (c) What is the Hall coefficient m3/c

A charged particle moves through a velocity selector at constant velocity. The velocity selector is configured with "crossed" electric and magnetic fields of magnitude E = 1.50 x 104 N/C and B = 0.4 T. %3D Hint a. What is the velocity of the charged particle? Velocity of the charged particle is x 104 m/s. b. When the electric field is turned off, the charged particle travels in a circular path of radius 6 mm, as it travels through the magnetic field (still at B = 0.4 T). What is the mass-to-charge ratio of the particle? Hint for (b) Mass-to-charge ratio of the part is kg/C. (Use the "E" notation to enter your answer in scientific notation. For example, to enter 3.14 × 10´ 12".) - 12 enter "3.14E-

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.

99. A velocity selector consists of electric and magnetic w fields described by the expressions E Bi, with B = 15.0 mT. Find the value of E such that a 750-eV electron moving in the negative x direction is undeflected. = Ek and B = %3D %3D %3D