# Force, Velocity, Mass, EMF, Current, Magnetic Field, and Torque

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1. A magnetic levitation train runs on two parallel rails, 1.20m apart. The rails each carry the same current, I = 1.00 Ã— 103 A, but in opposite directions. One section of rail is 20.0 m long. What is the magnitude and direction of the total force acting between the rails along one complete section?

Solution: Force per unit length between two large parallel current carrying conductors with a spacing r is given by : F/L = (Î¼o/4ÐŸ) 2I1I2/r. The force is attractive if the direction of current is same in both the conductor and repulsive if the direction is opposite.

In the present case: I1 = I2 = 1000 A, r = 1.2 m

Force per unit length between the rails = F/L = (Î¼o/4ÐŸ) 2I1I2/r = 10-7 x 2 x 1000 x 1000/1.2 = 0.167 N

Force on 20 m long section = F = 0.167 x 20 = 3.33 N

As the directions of currents are opposite, the force is repulsive.

2. In the velocity selector region of a mass spectrometer, an electric field of 2.34 Ã— 103 N/C is perpendicular to a magnetic field, B = 1.56 Ã— 10âˆ’2 T. A beam of positive ions (q = +e) then enters a region with the same B field, but no E field. If the ions are detected at a radius of curvature of r = 1.20m, what is the mass of the unknown ion?

Solution: Let us consider a positively charged particle enter the fields. It will experience a force of magnitude qE in the direction of E vector. Also the particle moving in the magnetic field experiences a force q(v X B) of magnitude qvBsinÎ¸.= qvB (as Î¸ = 90O). As a result of these forces, the particle will deviate from its straight line path. Only those particles are unaffected for which the two forces cancel out. Hence,

qvB = qE or v = E/B ........(1)

Particles which satisfy (1) continue to move in a straight line. Other particles deviate one way or the other depending upon whether qE > qvB or qvB > qE.

Hence, speed of the ions which continue to move straight = v = 2.34 Ã— 103/1.56 Ã— 10âˆ’2 = 1.5 x 105 m/s

The ions moving with speed 1.5 x 105 m/s enter the magnetic field (without electric field). The ions will start rotating in a circular path of radius r with a constant speed v. The force experienced by the ions is given by F = qvB = 1.6 x 10-19 x 1.5 x 105 x 1.56 Ã— 10âˆ’2 = 3.74 x 10-16 N

This force provides the centripetal force required for rotation of the particle in a circular path. Hence, mv2/r = 3.74 x 10-16

m = 3.74 x 10-16 x 1.2/(1.5 x 105)2 = 2 x 10-26 kg

3. A circular loop, 10.0 cm in diameter, rotates from Î¸ = 30 degrees to Î¸ ï€ = 50 degrees in 0.15 seconds, relative to a constant magnetic field of B = 0.800 T (see the figure below). Find the average emf induced in the loop during this rotation.

Solution: Flux linking with the loop = Ï† = B . A = BAcosÎ¸ where A = Area of the loop = ÐŸr2 = ÐŸ(0.05)2 = 7.85 x 10-3 m2

Flux Ï† = 0.8 x 7.85 x 10-3 cosÎ¸ = 6.28 x 10-3 cosÎ¸

As per Faraday's law of electromagnetic induction, EMF induced in the coil is given by:

e = - dÏ†/dt = 6.28 x 10-3 sinÎ¸ dÎ¸/dt .......(1) where dÎ¸/dt = Rate of change of angle Î¸ (in radians per sec)

Initial angle Î¸ = 30O = (ÐŸ/180) x 30 = 0.52 radians

Final angle Î¸ = 50O = (ÐŸ/180) x 50 = 0.87 radians

Time = 0.15 sec

Hence, dÎ¸/dt = (0.87 - 0.52)/0.15 = 2.33 rad/sec

Substituting in (1): e = 6.28 x 10-3 x 2.33 sinÎ¸ = 14.63x10-3sinÎ¸ Volts

0.87 rad

Average emf during Î¸ = 30O and Î¸ = 50O = eave = [14.63x10-3 âˆ«sinÎ¸ dÎ¸]/(0.87 - 0.52) =

0.52 rad

eave = [14.63x10-3(cos 0.52 rad - cos 0.87 rad)]/0.35 = [14.63x10-3(0.87 - 0.64)]/0.35 = 9.61 V

4. A piece of wire slides, without friction, along two ...

#### Solution Summary

The solution assists in providing answers for 15 physics questions, regarding force, velocity, mass, EMF, current, magnetic field, and torque.