At a particular point on Earth, the magnitude of the Earth's magnetic field is 6.50 × 10-5 T. At this point, the field through a small square loop is essentially uniform. The area of the loop is 10.0 cm². Whatis the magnitude of the magnetic flux (in T • m²) through the loop when the loop is at the following orientations?(a) the magnetic field is perpendicular to the plane of the loop6.5X2What is the definition of magnetic flux? How does it depend on magnetic field, area, and the angle between the field and the normal to the plane of the loop? Be careful with unit conversions. T · m²(b) the magnetic field and the normal to the plane of the loop are at a 53.0° angle2T·m²(c) the magnetic field and the normal to the plane of the loop are at a 90.0° angleT·m²

In the question, it is given. At a particular point on Earth, the magnitude of the Earth's magnetic…

At a particular point on Earth, the magnitude of the Earth's magnetic field is 6.50 x 10 T. At this point, the field through a small square loop is essentially uniform. The area of the loop is 10.0 cm². What is the magnitude of the magnetic flux (in T. m²) through the loop when the loop is at the following orientations? (a) the magnetic field is perpendicular to the plane of the loop 6.5 x What is the definition of magnetic flux? How does it depend on magnetic field, area, and the angle between the field and the normal to the plane of the loop? Be careful with unit conversions. T. m² (b) the magnetic field and the normal to the plane of the loop are at a 53.0° angle T-m² (c) the magnetic field and the normal to the plane of the loop are at a 90.0° angle T-m²

At a particular point on Earth, the magnitude of the Earth's magnetic field is 6.50 x 10 T. At this point, the field through a small square loop is essentially uniform. The area of the loop is 10.0 cm². What is the magnitude of the magnetic flux (in T. m²) through the loop when the loop is at the following orientations? (a) the magnetic field is perpendicular to the plane of the loop 6.5 x What is the definition of magnetic flux? How does it depend on magnetic field, area, and the angle between the field and the normal to the plane of the loop? Be careful with unit conversions. T. m² (b) the magnetic field and the normal to the plane of the loop are at a 53.0° angle T-m² (c) the magnetic field and the normal to the plane of the loop are at a 90.0° angle T-m²

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Magnetic Forces And Magnetic Fields

Section: Chapter Questions

Problem 63P

See similar textbooks

Similar questions

When the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0104 T. What is the radius of the loop?
Acircularcoiofwireofradius5.Ocmhas2Otums and carries a current of 2.0 A. The coil lies in a magnetic field of magnitude 0.50 T that is directed parallel to the plane of the coil. (a) What is the magnetic dipole moment of the coil? (b) What is the torque on the coil?
The accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius r1= 3.0 cm and outer radius r2= 5.0 cm. A 50-A current distributed uniformly over the cross-section flows into the page. Calculate the magnetic field at r = 2.0 cm. r = 4.0 cm. and r = 6.0 cm.
A magnetic field directed into the page changes with time according to B = 0.030 0t2 + 1.40, where B is in teslas and t is in seconds. The field has a circular cross section of radius R = 2.50 cm (see Fig. P23.28). When t = 3.00 s and r2 = 0.020 0 m, what are (a) the magnitude and (b) the direction of the electric field at point P2?
A long, solid, cylindrical conductor of radius 3.0 cm carries a current of 50 A distributed uniformly over its cross-section. Plot the magnetic field as a function of the radial distance r from the center of the conductor.
Two long coaxial copper tubes, each of length L, are connected to a battery of voltage V. The inner tube has inner radius o and outer radius b, and the outer tube has inner radius c and outer radius d. The tubes are then disconnected from the battery and rotated in the same direction at angular speed of radians per second about their common axis. Find the magnetic field (a) at a point inside the space enclosed by the inner tube r d. (Hint: Hunk of copper tubes as a capacitor and find the charge density based on the voltage applied, Q=VC, C=20LIn(c/b) .)
A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
Calculate the magnitude of the magnetic field at a point 25.0 cm from a long, thin conductor carrying a current of 2.00 A.
Figure CQ19.7 shows a coaxial cable carrying current I in its inner conductor and a return current of the same magnitude in the opposite direction in the outer conductor. The magnetic field strength at r = r0 is Find the ratio B/B0, at (a) r = 2r0 and (b) r = 4r0. Figure CQ19.7
A particle moving downward at a speed of 6.0106 m/s enters a uniform magnetic field that is horizontal and directed from east to west. (a) If the particle is deflected initially to the north in a circular arc, is its charge positive or negative? (b) If B = 0.25 T and the charge-to-mass ratio (q/m) of the particle is 40107 C/kg. what is ±e radius at the path? (c) What is the speed of the particle after c has moved in the field for 1.0105s ? for 2.0s?
How many turns must be wound on a flat, circular coil of radius 20 cm in order to produce a magnetic field of magnitude 4.0105 T at the center of the coil when the current through it is 0.85 A?
In Figure P22.43, the current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries a current I2 = 10.0 A. The dimensions in the figure are c = 0.100 m, a = 0.150 m, and = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. Figure P22.43 Problems 43 and 44.