The earth\'s magnetic field has an average value of 5 × 10 5 T. You may assume t

Question

The earth's magnetic field has an average value of 5

× 105

T. You may assume that this is a uniform magnetic field anywhere along the surface of the earth.

a) If a proton is traveling through earth's magnetic field, how fast would it need to be going in order to levitate? (Hint: The weight of the proton would need to be counteracted by the magnetic force on it.)

m/s

b) In what direction would the proton need to be traveling in order to levitate? (Hint: What is the compass direction for earth's magnetic field? What should be the direction of force on the proton for it to levitate?)

---Select--- North South East West Up Down

Explanation / Answer

(a) For proton to leviate, weight of proton must be balance by force due to magnetic field

Weight = mg

Force due to field B = qvB

where m = 1.67 x 10^-27 kg ; g = 9.81 m/s^2 ; q = 1.6 x 10^-19 C, v= ? B = 5 x 10^-T

qvB= mg

(1.67 x 10^-27)(9.81) = (1.6 x 10^-19)(v)(5 x 10^-5)

v = 2 x 10^-3 m/s

(b) Earth's magnetic field is towards North direction. Proton must leviate in upward direction.

Using Fleming's Left hand rule, Index finger points in direction of magnetic field(North) , Thumb points in direction of force(upwards), centre finger points in direction of velocity. Velocity is towards East

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