To practice Problem$-$Solving Strategy $27\mathrm{.}1$: Magnetic Forces.

A particle with mass $1\mathrm{.}81103$ kgkg and a charge of $1\mathrm{.}22108$ CC has, at a given instant, a velocity v$=(3\mathrm{.}00104$m$/$s$)$j$^$v$=(3\mathrm{.}00104$m$/$s$)$j$^.$What are the magnitude and direction of the particle$$s acceleration $$produced by a uniform magnetic field B$=(1\mathrm{.}63$T$)$i$^+(0\mathrm{.}980$T$)$j$^$B$=(1\mathrm{.}63$T$)$i$^+(0\mathrm{.}980$T$)$j$^?$

Problem$-$Solving Strategy $27\mathrm{.}1$: Magnetic Forces

IDENTIFY the relevant concepts:

The right$-$hand rule allows you to determine the magnetic force on a moving charged particle.

SET UP the problem using the following steps:

Draw the velocity vector v$$v$$ and magnetic field B$$B$$ with their tails together so that you can visualize the plane in which these two vectors lie.

Identify the angle $$ between the two vectors.

Identify the target variables. This may be the magnitude and direction of the force, the velocity, or the magnetic field.

EXECUTE the solution as follows:

Express the magnetic force using the equation F$=$qv$$B$$F$=$qv$$B$.$ The $$magnitude of the force is given by F$=$qvBsin$$F$=$qvBsin$.$

Remember thatF$$F$$ is perpendicular to the plane of the vectors v$$v$$ and B$$B$.$ The direction of v$$B$$v$$B$$ is determined by the righthand $$rule$.$ If qq is negative, the force is opposite to v$$B$$v$$B$.$

EVALUATE your answer:

Whenever you can, solve the problem in two ways. Verify that the results agree