Question

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases from 0.50 T to 1.50 T in 0.60 s. You may want to review (Pages 825 - 829) . For help with math skills, you may want to review: Rearrangement of Algebraic Expressions For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Electromagnetic induction. Part APart complete What is the induced emf in the coil?

Answer 1

Answer:

The induced emf is  $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

     The number of turn is  $N = 100$

      The diameter of the coil is  $d = 2.0 cm = 2.0 *10^{-2} m$

      The uniform magnetic at initial is $B_i = 0.50 T$

      The uniform magnetic at initial is $B_f = 1.50 T$

      The time taken is $t = 0.60s$

      The angle the magnetic field makes with vertical is  $\theta = 60^o$

Generally induced emf is mathematically represented as

     $\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change  magnetic flux

Magnetic flux is mathematically represented as

       $\O = \= B \cdot \= A$

          $= BA cos \theta$

Substituting this above  

       $\epsilon = -N \frac{d (BA cos \theta)}{dt}$

      $\epsilon = -N A \frac{d (B cos \theta)}{dt}$

  Where B is the magnetic field and A is the area which is mathematically evaluated as

        $A = \frac{\pi d^2}{4}$

Substituting values

        $A = \frac{\pi (2.0 *10^{-2})^2}{4}$

           $A= 3.142*10^{-4}m^2$

From the equation of emf

          $\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

       So

             $|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

            $|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

                $|\epsilon|=0.0261 V$