Question

Early in 1981 the Francis Bitter National Magnet Laboratory at M.I.T. commenced operation of a 3.3 cm diameter cylindrical magnet, which produces a 30 T field, then the world's largest steady-state field. The field can be varied sinusoidally between the limits of 29.6 and 30.0 T at a frequency of 15 Hz. When this is done, what is the maximum value of the induced electric field at a radial distance of 1.4 cm from the axis

Answer 1

Answer:

The maximum electric field $E_{max}= 0.132V/m$

Explanation:

From the question we are told that

       The diameter is  $d = 3.3 cm = \frac{3.3}{100} = 0.033m$

         The magnetic field  of the cylinder is $B = 30 T$

           The frequency is  $f = 15Hz$

            The radial distance is  $d_r = 1.4cm = \frac{1.4}{100} = 0.014m$

This magnetic field can be represented mathematically as

        $B(t) = B_i + B_1sin (wt + \o_i)$

 The initial magnetic field is the average between the variation of the magnetic field which is represented as

          $B_i = \frac{30 + 29.6}{2}$

               $= 29.8T$

Then $B_1$ is the amplitude of the  resultant  field is mathematically evaluated as

                 $B_1 = \frac{30.0 - 29.6}{2}$

                       $= 0.200T$

The electric field induced can be represented mathematically as

        $E = \frac{1}{2} [\frac{dB }{dt} ]d_r$

            $= \frac{d_r}{2} \frac{d}{dt} (B_i + B_1 sin (wt + \o_o))$

            $= \frac{1}{2} (B wr cos (wt + \o_o))$

At maximum electric field  $cos (wt + \o_o) = 1$

        $E_{max} = \frac{1}{2} B_1 wd_r$

         $E_{max} = \frac{1}{2} B_1 2 \pi f d_r$

                  $= \frac{1}{2} (0.200) (2 \pi (15 ))(0.014)$

                 $E_{max}= 0.132V/m$