Una espira plana de alambre formada por una sola vuelta de 8.00 cm2 de área de sección transversal es perpendicular a un campo magnético que aumenta uniformemente de magnitud de 0.500 T a 2.50 T en 1.00 s. ¿Cuál es la corriente inducida resultante si la espira tiene una resistencia de 2Ω

Explanation:

Given that,

Cross-sectional area, A = 8 cm² = 0.0008 m²

The initial magnetic field,

The final magnetic field,

Time, t = 1 s

The resistance of the loop, R = 2 Ω

We need to find the induced current. Due to the change in the magnetic field, emf will include in the loop. The induced emf is given by :

Let I is the induced current. Using Ohm's law,

V = IR

So, the induced current in the loop is .

Hello, Here is your answer ..

- V = 4/3*pi*(d/2)^3. In mks, then cm-g-s units, ρ = m/V = (0.004/6.02214179E+23) / (4/3*pi*(1E-10/2)^3) = 12685.582 kg/m^3 = 12.685582 .

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