Question#2 A particle with mass 1.81x10"kg and a charge of 1.22x10"-Chas, at a given instant, a velocity v = (3.00 x 10* m/s)/. What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field & = (1.63 1)i + 10.980 1)j? Problem-Solving Strategy 27.1: Magnetic Forces IDENTIFY the relevant concepts The right hand rule alows you to determine the magnetic force on a moving charged particle. SET UP the problem using the following steps. 1. Draw the velocity vector vv_vec and magnetic field BB_vec with their tails together so that you can visualize the plane in which these two vectors le. 2 Identify the angle ophi between the two vectors. 3. Identify the target variables. This may be the magnitude and direction of the force, the velocity, or the magnetic field. EXECUTE the solution as folows 1. Express the magnetic force using the equation F = qv * B. The magnitude of the force is given by F = qubeing. 2 Remember thatF_vec is perpendicular to the plane of the vectors iv_ver and JIB_ ver. The direction of 1 = $ is determined by the nighthand rule. If qq is negative, the force is opposite to p x B. EVALUATE your answer Whenever you can solve the problem in two ways. Verify that the results agree