A dust particle has mass of 7.2 times l0^-10 kg. The Earth\'s magnetic field at

Question

A dust particle has mass of 7.2 times l0^-10 kg. The Earth's magnetic field at the location of the particle is 3.5 times 10^-5 T. The charge on the particle is 3.6 times 10^-11 C. a) Compute the weight w in newtons of the particle. b) Compute v, the speed of the particle if its magnetic force balances its weight. c) Compute E, the electric field strength required for a velocity selector that allows the charged dust particle to pass through undeflected. (Assume that the magnetic field is the same as above.) d) If the plate separation of the velocity selector is 7.5 mm, compute the potential difference across its plates. e) If a solenoid with 1547 turns per meter produces the same magnetic field as the Earth given above, then compute the current through the coils.

Explanation / Answer

a)
The weight of the particle will be the acceleration due to gravity times the mass of the particle.
W = mg = 7.2*10^-10*9.81 = 7.0632*10^-9 N

b) The magnetic force is given by q v B where B is the magnetic field strength, q is the charge and v is the velocity.
Since it is balanced by the weight,
q v B = W
v = W/qB = 7.0632*10^-9/(3.6*10^-11*3.5*10^-5) = 5605714.29 m/s.

c) The electric field will be the velocity times the magnetic field strength.
i.e. E = vB or E = 5605714.29 * 3.5*10^-5 = 196.2 V/m.

d) The potential difference across its plate = Electric field strength times the distance (separation) = E*d = 196.2*7.5/1000 = 1.4715 V

e)
The magnetic field due to solenoid = mu n I = 4pi*1537*I = 3.5*10^-5
I = 3.5*10^-5/(4pi*1537) = 1.81*10^-9 A.
I = 1.91 nA.

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