Question

A square coil with a side length of 16.0 cm and 29 turns is positioned in a region with a horizontally directed, spatially uniform magnetic field of 83.0 mT and set to rotate about a vertical axis with an angular speed of 1.20 ✕ 102 rev/min.
(a) What is the maximum emf induced in the spinning coil by this field?
___V
(b) What is the angle between the plane of the coil and the direction of the field when the maximum induced emf occurs? (Enter the angle with the smallest possible magnitude.)
___°

Answer 1

Answer:

A

  $\epsilon_{max} = 0.774 \ V$

B

$wt = 0^o$

Explanation:

From the question we are told that

   The  length of the side is  $l = 16.0 \ cm = 0.16 \ m$

    The  number of turns is  $N = 29 \ turns$

    The  magnetic field is  $B = 83.0 mT = 83 *10^{-3} \ T$

    The  angular speed is  $w = 1.20 * 10^2 rev/min = \frac{1.20 *10^2 * 2\pi}{60 } = 12.6 \ rad/s$

Generally  the area is  $A = l^2$

Generally the induced emf is mathematically represented as

         $\epsilon = N * w * B * A * cos(wt)$

At maximum  $cos(wt) = 1$

    So

       $\epsilon_{max} = N * w * B * A$

       $\epsilon_{max} = 29 * 12.6 * 83*10^{-3}* (l^2)$

=>   $\epsilon_{max} = 29 * 12.6 * 83*10^{-3}* ((0.16)^2)$

=>  $\epsilon_{max} = 0.774 \ V$

At maximum emf

      $cos (wt) = 1$

=>   $(wt) = cos^{-1} (1)$

=>   $wt = 0^o$