Question

A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axis of the coil is parallel to the field What is the emf of the coil?

Answer 1

Answer:

The induced voltage is  $\epsilon = 4.53 \ V$

Explanation:

From the question we are told that

    The  number of turns is  $N = 1300 \ turns$

     The diameter is  $d = 2.2 \ cm =0.022 \ m$

     The  initial magnetic field is  $B_i = 0.11 \ T$

     The final magnetic field is  $B_f = 0 \ T$

    The time taken is  $t = 12 \ ms = 12*10^{-3} \ s$

   

The radius is mathematically evaluated as

         $r = \frac{d}{2}$

substituting values

         $r = \frac{0.022}{2}$

         $r = 0.011 \ m$

Generally the induced emf  is mathematically represented as

      $\epsilon = - N * \frac{d\phi}{dt}$

Where  $d\phi$ is the change in magnetic flux of the wire which is mathematically represented as

      $d \phi = dB* A * cos \theta$

=>  $d \phi = (B_f - B_i )* A * cos \theta$

Here  $\theta = 0$

since the axis of the coil is parallel to the field

    Where A  is the cross-sectional area of the coil which is mathematically represented as

      $A = \pi * r^2$

       $A = 3.142 * 0.011^2$

      $A = 3.80*10^{-4} \ m^2$

So the induced emf

        $\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}$   Here we substituted the values of  $d \phi$

       $\epsilon = 4.53 \ V$

Answer 2

The emf induced in the coil at the given magnetic field strength is 4.53 V.

The given parameters;

• number of turns, N = 1300 turns
• diameter of the coil, d = 2.2 cm
• initial magnetic field, B₁ = 0.11 T
• final magnetic field, B₂ 0
• time, t = 12 ms

The area of the coil is calculated as follows;

$A = \pi r^2 = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.022^2}{4} = 0.00038 \ m^2$

The emf induced in the coil is calculated as follows;

$emf = -N\frac{d\phi}{dt} \\\\emf = N (\frac{\phi_1 - \phi_2}{t} )\\\\emf = N(\frac{AB_1 - AB_2}{t} )\\\\emf = NA(\frac{B_1 - B_2}{t} )\\\\emf = 1300 \times 0.00038 (\frac{0.11 - 0}{12 \times 10^{-3}} )\\\\emf = 4.53 \ V$

Thus, the emf induced in the coil at the given magnetic field strength is 4.53 V.

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