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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 × 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 = (μo/4П) 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 = (μo/4П) 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 × 103 N/C is perpendicular to a magnetic field, B = 1.56 × 10−2 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 qvBsinθ.= qvB (as θ = 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 × 103/1.56 × 10−2 = 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 × 10−2 = 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 θ = 30 degrees to θ = 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 = φ = B . A = BAcosθ where A = Area of the loop = Пr2 = П(0.05)2 = 7.85 x 10-3 m2

Flux φ = 0.8 x 7.85 x 10-3 cosθ = 6.28 x 10-3 cosθ

As per Faraday's law of electromagnetic induction, EMF induced in the coil is given by:
e = - dφ/dt = 6.28 x 10-3 sinθ dθ/dt .......(1) where dθ/dt = Rate of change of angle θ (in radians per sec)

Initial angle θ = 30O = (П/180) x 30 = 0.52 radians

Final angle θ = 50O = (П/180) x 50 = 0.87 radians

Time = 0.15 sec

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

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

0.87 rad
Average emf during θ = 30O and θ = 50O = eave = [14.63x10-3 ∫sinθ dθ]/(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.

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