Share
Explore BrainMass

Electromagnet, magnetic force, positive ion, parallel wire, coil

A laboratory electromagnet produces a magnetic field of magnitude 1.50 T. A proton moves through this field with a speed of 6.00 x 10^6 m/s. (a) Find the magnitude of the maximum magnetic force that could be exerted on the proton. (b) What is the magnitude of the maximum acceleration of the proton? (c) Would the field exert the same magnetic force on an electron moving through the field with the same speed? Would the electron undergo the same acceleration? Explain.

A wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire for the following angles between the magnetic field and the current, (a) 60.0o (b) 90.0o, and (c) 120.0o.

A singly charged positive ion has a mass of 2.50 x 10^-26 kg. After being accelerated through a potential difference of 250 V, the ion enters a magnetic field of 0.500 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Two wires carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field (a) at a point midway between the wires; (b) at point P1, 10.0 cm to the right of the wire on the right; and (c) at point P2, 20.0 cm to the left of the wire on the left

Two parallel wires are separated by 6.00 cm, each carrying 3.0 A of current in same direction. What is the magnitude of the force per unit length between the wires? Is the force attractive or repulsive?

A 30-turn circular coil of length 6.0 produces a magnetic field of magnitude 2.0 mT at its center. Find the current in the loop.

A uniform magnetic field of magnitude 0.50 T is directed perpendicular to the plane of a rectangular loop having dimensions 8.0 cm by 12 cm. Find the magnetic flux through the loop.

A circular loop of radius 12.0 cm is placed in a uniform magnetic field (a) If the field is directed perpendicular to the plane of the loop and the magnetic flux through the loop is 8.00 x 10^-3 T x m^2, what is the strength of the magnetic field? (b) If the magnetic field is directed parallel to the plane of the loop, what is the magnetic flux through the loop?

A wire loop of radius 0.30m lies so that an external magnetic field of strength +0.30T is perpendicular to the loop. The field changes to −0.20T in 1.5 s. Find the magnitude of the average induced emf in the loop during this time.

A pick-up truck has a width of 79.8 in. If it is traveling north at 37 m/s through a magnetic field with vertical component of 35µT, what magnitude emf is induced between the driver and passenger sides of the truck?

Solution Preview

Please refer to the attachment.

1. The Lorentz force experienced by a charge q moving with a velocity v in a magnetic field B is given by: F = q(v X B) [where v X B is the cross product of velocity and magnetic field vectors]

a) Magnitude of force F is given by: F = qvBsinθ where θ is the angle between v and B vectors
Maximum magnitude of F is for θ = 90O. Hence, Fmax = qvB

Substituting q = 1.6 x 10-19 C, v = 6 x 106 m/s, B = 1.5 T, we get:
Fmax = 1.6 x 10-19 x 6 x 106 x 1.5 = 14.4 x 10-13 N

b) Maximum acceleration = Fmax/mproton where mproton = mass of proton = 1.672x10-27 kg

Maximum acceleration of proton = 14.4 x 10-13/1.672x10-27 = 8.61x1014 m/s2

c) As the charge on the electron is same as that on the proton, the maximum force experienced by the electron moving with the same speed will be same as that for proton. However, electron's acceleration will be much higher on account of it's lesser mass.

Maximum acceleration of electron = 14.4 x 10-13/9.11x10-31 = 1.58x1018 m/s2

2. Force experienced by a conductor of length l carrying current I in a uniform magnetic field B is given by: F = I (l X B)

B

θ I

Magnitude of force F = BIl sinθ where θ is the angle between ...

Solution Summary

The expert examines electromagnet, magnetic forces, positive ion, parallel wire and coils in a laboratory.

$2.19