Question

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^v→=(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^B→=(1.63T)i^+(0.980T)j^?

Answer 1

Answer:

The magnitude of the acceleration  is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle $m = 1.8*10^{-3} kg$

         The charge on the particle is $q = 1.22*10^{-8}C$

         The velocity is $\= v = (3.0*10^4 m/s ) j$

        The the magnetic field is  $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced  a force which is mathematically represented as

         

                    $F = q (\= v * \= B)$

    Substituting value

         $F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

            $= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

            $= (-5.966*10^4 N) \r k$

Note :

           $i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j$

Now force is also mathematically represented as

        $F = ma$

Making a the subject

      $a = \frac{F}{m}$

   Substituting values

     $a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

        $= (-0.33m/s^2)\r k$

        $= 0.33m/s^2 * (- \r k)$