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A particle with a mass of 2.00×10−4 kg carries a negative charge of - 3.80×10−8 C . The particle is given an initial horizontal velocity that is due north and has a magnitude of 4.21×104 m/s . A) What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

B) What is the direction of the minimum magnetic field? (West, East, North, South)

Respuesta :

lavanyaande

Answer:

A. The Magnitude is 1.2251 TESLA which keeps the particle moving in the earth's gravitational field in the same horizontal, northward direction.

B. The direction of “minimum magnetic field” is East.

Explanation:

(A) To calculate Magnitude:  

From gravitational force (mg) = magnetic force (qvB).

qvB = mg

Given that,

[tex]q=3.80 \times 10^{-8} \mathrm{C}[/tex]

[tex]v=4.21 \times 10^{4}[/tex]

[tex]\mathrm{m}=2.00 \times 10^{-4} \mathrm{kg}[/tex]

[tex]\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}[/tex]

Substitute the given values in the formula,

[tex]3.80 \times 10^{-8} \times 4.21 \times 10^{4} \times B=2.00 \times 10^{-4} \times 9.8[/tex]

[tex]15.998 \times 10^{-4} \times \mathrm{B}=19.6 \times 10^{-4}[/tex]

[tex]\mathrm{B}=\frac{19.6}{15.998}[/tex]

B = 1.2251 TESLA

(B) Direction of minimum magnetic field:

The direction is east. Force, F = q(vXB) = mg.  

Applying “Fleming's left hand rule” for vector product we know that the “negative charge” moving towards “north” establishes “conventional current” towards “south” the “magnetic field” has to be “eastwards” so, that the force due to applied “magnetic field” acts in the reverse direction of the “gravity of earth”. The “earth's gravitational field” will be ineffective in the commencement because the charge moves same the direction of “magnetic field” due to earth. To balance “gravity” the “magnetic force” has to point upward and have the “same magnitude” as the “gravitational force”. From “right-hand rule” take that the particle has "negative" charge then the field has to point to the "East" to create upward force.

onyebuchinnaji

(a) The minimum magnetic field of the particle is 1.23 T.

(b) The direction of the minimum magnetic field is South.

Minimum magnetic field of the particle

The minimum magnetic field of the particle is calculated as follows;

mg = qvB

B = mg/qv

Where;

  • m is mass of the particle
  • q is charge of the particle
  • v is the velocity
  • B is the magnetic field

[tex]B = \frac{ 2\times 10^{-4} \times 9.8}{(3.8 \times 10^{-8} ) \times 4.21 \times 10^4} \\\\B = 1.23 \ T[/tex]

Direction of the magnetic field

The direction of the magnetic field will be in opposite direction to the charge. Thus the direction of the minimum magnetic field is South.

Learn more about direction of magnetic field here: https://brainly.com/question/18649497