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A 325-turn circular-loop coil 9.4 cm in diameter is initially aligned so that its axis is parallel to Earth’s magnetic field. In 2.48 ms the coil is flipped so that its axis is perpendicular to Earth’s magnetic field. If an average voltage of 0.17 V is thereby induced in the coil, what is the value of Earth’s magnetic field at that location? Answer in units of µT.

Respuesta :

Answer:

[tex]B = 187\ \mu T[/tex]

Explanation:

Given,

Number of turns, N = 325

Diameter of coil, d = 9.4 cm

time,t = 2.48 m s

average voltage,ε = 0.17 V

Earth magnetic field, B = ?

We know that

[tex]\epsilon = \dfrac{NBA}{t}[/tex]

[tex]0.17= \dfrac{325\times B \times \pi (0.047)^2}{2.48\times 10^{-3}}[/tex]

[tex]B = 187 \times 10^{-6}\ T[/tex]

[tex]B = 187\ \mu T[/tex]

Magnetic field of the earth is equal to [tex]B = 187\ \mu T[/tex]

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