A coil has 170 turns enclosing an area of 14.0 cm2cm2 . In a physics laboratory experiment, the coil is rotated during the time interval 0.044 ss from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 6.50×10−5 TT .Part APart completeWhat is the magnitude |Φinitial||Φinitial| of the magnetic flux through one turn of the coil before it is rotated?Express your answer in webers.View Available Hint(s)for Part A|Φinitial||Φinitial| =9.10×10−8 WbWb SubmitPrevious Answers CorrectPart BWhat is the magnitude |Φfinal||Φfinal| of the magnetic flux through one turn of the coil after it is rotated?
Question
A coil has 170 turns enclosing an area of 14.0 cm2cm2 . In a physics laboratory experiment, the coil is rotated during the time interval 0.044 ss from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 6.50×10−5 TT .Part APart completeWhat is the magnitude |Φinitial||Φinitial| of the magnetic flux through one turn of the coil before it is rotated?Express your answer in webers.View Available Hint(s)for Part A|Φinitial||Φinitial| =9.10×10−8 WbWb SubmitPrevious Answers CorrectPart BWhat is the magnitude |Φfinal||Φfinal| of the magnetic flux through one turn of the coil after it is rotated?
Solution
Part A: The magnetic flux (Φ) through a surface is given by the dot product of the magnetic field (B) and the area (A) vector. When the coil is perpendicular to the magnetic field, the angle between the magnetic field and the area vector is 0 degrees. The dot product simplifies to BAcos(0), and since the cosine of 0 degrees is 1, the initial magnetic flux is just the product of the magnetic field and the area.
Given: B = 6.50×10−5 T (Tesla) A = 14.0 cm2 = 14.0 x 10^-4 m2 (converted from cm2 to m2)
So, Φinitial = B * A = 6.50×10−5 T * 14.0 x 10^-4 m2 = 9.10 x 10^-8 Wb (Weber)
Part B: When the coil is parallel to the magnetic field, the angle between the magnetic field and the area vector is 90 degrees. The dot product simplifies to BAcos(90), and since the cosine of 90 degrees is 0, the final magnetic flux is 0.
So, Φfinal = 0 Wb.
Similar Questions
A coil with 155 turns of diameter 22.0 cm rotates at 1250 rpm in Earth’s magnetic field.Calculate the maximum induced emf, if Earth’s magnetic field is 3.80 x 10-5 T. a.29.3 mV.b.2.93 mV.c.0.00293 V.d.0.293 V
A coil of wire enclosing an area 100 cm2 is placed with its plane making an angle 60º with the magnetic field ofstrength 10-1T. What is the flux through the coil? If magnetic field is reduced to zero in 10-3 s, then find the inducedemf
A thin coil has 14 rectangular turns of wire. When a current of 1.00 A runs through the coil, there is a total flux of 0.001 T-m2 enclosed by one turn of the coil (note that the inductance is the proportionality constant between current and magnetic flux). What is the total inductance in Henries?
A 26-turn circular coil of wire has radius 0.4 m. It is placed with its axis along the direction of the Earth's magnetic field of 58.0 µT and then in 0.200 seconds is flipped 180°. An average emf of what magnitude is generated in the coil? (units of V)Answer
What average emf (in volts) is induced, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 8 ms?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.