Question

A time-dependent but otherwise uniform magnetic field of magnitude B0(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 5.5 T, but it is decreasing at a rate of 24.5 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. What is the direction of acceleration of a proton placed in at 1.5 cm from the center?

Answer 1

Answer:

The acceleration is   $a = 3.45*10^{3} m/s^2$

Explanation:

 From the question we are told that

         The radius is  $d = 6.5 cm = \frac{6.5}{100} = 0.065 m$

           The magnitude of the magnetic field is  $B = 5.5 T$

           The rate at which it decreases is  $\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s$

             The distance from the center of field is  $r = 1.5 cm = \frac{1.5}{100} = 0.015m$

  According to Faraday's law

          $\epsilon = - \frac{d \o}{dt}$

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

 Where  the magnetic flux $\o = B* A$

             E is the electric field  

             dl is a unit length

 So

         $\int\limits {E} \, dl = - \frac{d}{dt} (B*A)$

         ${E} l = - \frac{d}{dt} (B*A)$

Now $l$ is the circumference of the circular loop formed by the magnetic field and it mathematically represented as  $l = 2\pi r$

A is the area  of the circular loop formed by the magnetic field and it 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 values  

    $E = \frac{0.015}{2} (24*10^{-4})$

         $E = 3.6*10^{-5} V/m$

The negative signify the negative which is counterclockwise

 

  The force acting on the proton is mathematically represented as

                       $F_p = ma$

        Also       $F_p = q E$

So

           $ma = qE$

 Where m is the mass of the the proton which has a value of  $m = 1.67 *10^{-27} kg$

 $q = 1.602 *10^{-19} C$

     So

            $a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}$

               $a = 3.45*10^{3} m/s^2$