A circular coil of 8 turns and a diameterof 30.0 cm is oriented in a vertical pl

Question

A circular coil of 8 turns and a diameterof 30.0 cm is oriented in a vertical plane with its axisperpendicular to the horizontal component of the Earth's magneticfield. A horizontal compass placed at the center of the coil ismade to deflect 45.0° from magnetic north by a current of0.510 A in the coil. (a) What is the horizontal component of theEarth's magnetic field?
1 µT
(b) The current in the coil is switched off. A "dip needle" is amagnetic compass mounted so that it can rotate in a verticalnorth-south plane. At this location, a dip needle makes an angle of18.0° from the vertical. What is thetotal magnitude of the Earth's magnetic field at this location?
2 µT (a) What is the horizontal component of theEarth's magnetic field?
1 µT
(b) The current in the coil is switched off. A "dip needle" is amagnetic compass mounted so that it can rotate in a verticalnorth-south plane. At this location, a dip needle makes an angle of18.0° from the vertical. What is thetotal magnitude of the Earth's magnetic field at this location?
2 µT

Explanation / Answer

Given    Number of turns N = 8    radius r = 0.15m    Current I = 0.510 A    angle ? = 45 µo = 1.25663706 × 10^-6 m kg s^-2 A^-2 ---------------------------------------------------------- a) Magnitic field due to the circular coil is                B' = ( µoNI ) /( 2R)     the horizontal component of the Earth's magneticfield is            B = B' /tan45             = ( µoNI ) /( 2R tan45) b) the total magnitude of the Earth's magnetic field atthis location is            B = B' /sin18.0o                = ( µoNI ) /( 2R sin18.0o ) solve to get

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