Question

A time-dependent but otherwise uniform magnetic field of magnitude Bo(t) is confined in a cylindrical region of radius 6.5 cm. Initially the magnetic field in the region is pointed out of the page and has a magnitude of 3.5 T, but it is decreasing at a rate of 29.5 G/s. Due to the changing magnetic field, an electric field will be induced in this space which causes the acceleration of charges in the region.
Required:
What is the direction of acceleration of a proton placed in at the point 1.5 cm from the center?

Answer 1

Answer:

The acceleration is   $a = 9197.11 m/s^2$

Explanation:

     From the question we are told that

        The radius of the cylinder is $r = 6.5 \ cm = \frac{6.5}{100} = 0.065 m$

         The magnitude of the magnetic field is  $B_i = 3.5T$

         The rate of decrease is  $\frac{dB}{dt} = 29.5 \ G/s = \frac{29.5}{10000} = 29.5 * 10 ^{-4} T/s$

        The distance from the center is $D = 1.5\ cm = \frac{1.5}{10}=0.15m$

Faraday's law of  induction states mathematically that

       $\epsilon = \frac{d \o}{dt}$

       $\epsilon = \int\limits {E} \, dl$

Where $\epsilon$ is the induced emf

            E is the magnitude of  the  electric field

            $dl$ i the change in length

         $d \o$ is the change in magnetic flux which is mathematically represented as

         $\o = BA$

substituting this into the above equation

        $\int\limits {E} \, dl = \frac{d(BA)}{dt }$

         $E l = A \frac{dB}{dt}$

Where l is the circumference of the circular loop formed in the cylinder  which is mathematically represented as

          $l = 2\pi r$

And A is the area of the circular loop formed which is mathematically represented as

          $A = \pi r^2$

    So

        $E (2 \pi r ) = (\pi r^2 ) \frac{dB}{dt}$

           $E = \frac{r}{2} \frac{dB}{dt}$

Substituting value

           $E = \frac{0.065}{2} * 29.5 *10^{-4}$

                $= 9.588*10^{-5} \ V/m$

Generally acceleration is mathematically represented as

                 $a = \frac{F}{m}$

Now F is the electric force which is mathematically represented as

              $F = qE$

    Substituting this into the question

              $a = \frac{qE }{m }$

Where q is the  charge on the proton with a constant value of  

             $q = 1.60 *10^ {-19 } C$

     and  m is the mass of the proton with a constant value of

               $m = 1.67 *10^{-31} kg$

Substituting values

               $a = \frac{1.60*10^{-19} * 9.588 *10^{-5}}{1.67 *10^{-27}}$

                 $a = 9197.11 m/s^2$