A charged particle (q) moves in the homogeneous magnetic field B vector = B_0 i

Question

A charged particle (q) moves in the homogeneous magnetic field B vector = B_0 i vector and in the homogeneous electric field E vector = E_0j vector. The particle has mass m: at time t = 0, its position is r vector = 0 bar, and its velocity is v vector = v_0i vector. For time t = 0, find the magnetic force exerted on the particle in this magnetic field. Find the time dependence of the particle velocity v vector(t) = v_x(t) j vector + v_y(t)j vector + v_z(t) k vector = ?, and its position in space f(t) = x(t)i vector + y(t) j vector + z(t) k vector = ?. The system shown in Figure below is in the homogeneous magnetic field B = 0.1 T. The distance between two conducting vertical is t = 10 cm. A rod having a mass M = 10 g start moving in a gravitational field (initial velocity is zero), and it is always in electrical contact with the rails. the emf of the battery is epsilon = 12 V, and the resistance of the resistor is R = 4 Om, Two forces are exerted on the rod: one is the gravity force (F vector = mg bar), and the second is the force due to the magnetic field (F_A = |B|, its direction is given by the corresponding rule). At the beginning (v = 0), the current in the circuit is determined by the emf of the battery only. Under the action of these two forces, the rod starts moving, and once it acquires some velocity, the induction emf appears and changes the current in the circuit and the magnitude of the magnetic force. Eventually, the velocity of the rod reaches some magnitude at which the gravity force is equal to the magnetic force and, after that, there is no velocity change any more (because the acceleration of the rod is zero). Please, find this velocity of the rod. Find the magnitude of this velocity in the case when emf of the battery epsilon = 0 v. What is the magnitude of this velocity if also the resistance R is zero?

Explanation / Answer

From the given question,

When a conductor of length(l) moves with velocity(v) in a perpendicular magnetic field(B), emf(E) is induced across the ends of conductor.

E=Blv

B=0.1 T

l=10cm=0.1 m

M=10g=0.01 kg

Emf of battery(Eb)=12 V

Resistance(R)=4 Ohm

For steady velocity,

BIl=mg

B[(Blv-Eb)/r]l=mg

0.1[(0.1*0.1*v-12)/4]0.1=0.01(9.8)

v=5120 m/s

When Eb=0,

B[(Blv-0)/r]l=mg

0.1[(0.1*0.1*v-12)/4]0.1=0.01(9.8)

v=3920 m/s

v=1200 m/s

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