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aleksandrvk [35]
3 years ago
11

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^?
Physics
1 answer:
Roman55 [17]3 years ago
6 0

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)

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