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Sindrei [870]
4 years ago
5

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

d decreases to zero over a time of 0.10 sec. What is the induced emf?a. 0.21 Vb. 7.1 Vc. 0.01 Vd. 0.2.25 V
Physics
1 answer:
lana66690 [7]4 years ago
6 0

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

\epsilon = - d\frac{\phi_B}{dt}

\epsilon = -NA\frac{dB}{dt}

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}

\epsilon = 0.212V

\epsilon = 0.21V

Therefore the correct answer is A.

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The magnitude of the acceleration  is a = 0.33 m/s^2

The direction is - \r k i.e the negative direction of the z-axis

Explanation:

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       The mass of the particle m = 1.8*10^{-3} kg

         The charge on the particle is q = 1.22*10^{-8}C

         The velocity is \= v = (3.0*10^4 m/s ) j

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                    F = q (\= v * \= B)

    Substituting value

         F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \  ( 1.63 \r i  + 0.980 \r j )T)

            = 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \  X \ \  \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \  \r  j))

            = (-5.966*10^4 N) \r k

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           i \ \ X \ \ j = k \\\\j \ \  X  \ \ k = i\\\\k  \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \  X  \ \ j = -i\\\\i  \ \ X \ \ k = - j\\

Now force is also mathematically represented as

        F = ma

Making a the subject

      a = \frac{F}{m}

   Substituting values

     a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}

        = (-0.33m/s^2)\r k

        = 0.33m/s^2 * (- \r k)

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