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I am Lyosha [343]
3 years ago
6

In a certain region the average horizontal component of Earth's magnetic field is 17 μT, and the average inclination or "dip" is

79°. What is the corresponding magnitude of Earth's magnetic field in microteslas?
Physics
1 answer:
Rufina [12.5K]3 years ago
6 0

Answer:

the Earth magnetic field in that region is about 89.09 \muT

Explanation:

If the magnetic field has an inclination of 79^o, and its horizontal component is 17 microTesla, then we can calculate the magnitude of the full magnetic field via the cosine function. Notice that the decomposition of a vector in horizontal and perpendicular components originates a right angle triangle where the vector is the hypotenuse of the triangle.

cos(\theta)=\frac{adjacent}{hypotenuse}\\cos(79^o)=\frac{17}{B} \\B=\frac{17}{cos(79^o)}\,\mu T  \\B=89.09\,\,\mu T

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A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

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time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

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