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Semmy [17]
3 years ago
10

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

diameter and has 1500 turns. When turned on, the current in the solenoid is increases linearly to 20 A in 1 second. What is the induced emf in the ring
Physics
1 answer:
Troyanec [42]3 years ago
3 0

Answer:

The value is  \epsilon =  3.84 *10^{-5} \  V

Explanation:

From the question we are told that

  The diameter of the ring is  d =  18 \ mm  =  0.018 \  m

   The length of the solenoid is l = 25 \ cm  =  0.25 \ m

   The diameter of the solenoid is  D = 5.0 \ cm  = 0.05 \ m

    The number of turns is  N = 1500

   The change in  current in the solenoid is   \Delta  I   = 20 \ A

   The time taken is  \Delta  t  = 1 \ s

Generally the radius of the ring is  

     r = \frac{d}{2}

=>  r = \frac{0.018 }{2}

=>  r = 0.009 \ m

Generally the area of the ring is mathematically represented as  

      A = \pi r^2

=>   A = 3.142 *  0.009^2    

=>   A = 2.545 *10^{-4}\ m^2

Generally the induced emf is mathematically represented as

       \epsilon  =  A * \frac{dB}{dt}

Here    

         \frac{dB }{dt} =  \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}

Here  \mu_o is the permeability of free space with value  

         \mu_o =  4\pi *10^{-7} \ N/A^2

So  

     \frac{dB }{dt} =   4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}

=>  \frac{dB }{dt} =   0.150816\  T/s

So

     \epsilon =   0.150816 *  2.545 *10^{-4}

=>   \epsilon =  3.84 *10^{-5} \  V

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