Examkin

Soru

Zorluk: OrtaMagnetic Force and Electromagnetism

An electron carrying a charge of magnitude 1.6×10−19 C1.6 \times 10^{-19}\text{ C} moves with a velocity of 4.0×106 m s−14.0 \times 10^6\text{ m s}^{-1} at an angle of 30∘30^\circ relative to a uniform magnetic field. If the magnetic force acting on the electron is 3.2×10−13 N3.2 \times 10^{-13}\text{ N}, what is the magnetic flux density of the field?

  1. A
    0.5 T0.5\text{ T}
  2. 1.0 T1.0\text{ T}Cevap
  3. C
    2.0 T2.0\text{ T}
  4. D
    0.25 T0.25\text{ T}

Cevap

The magnetic flux density of the field is 1.0 T1.0\text{ T}.
Applying the Lorentz force equation for a charged particle F=qvBsin⁡θF = qvB\sin\theta, substituting the given values yields 3.2×10−13=(1.6×10−19)(4.0×106)Bsin⁡(30∘)3.2 \times 10^{-13} = (1.6 \times 10^{-19})(4.0 \times 10^6) B \sin(30^\circ). Solving for BB gives B=1.0 TB = 1.0\text{ T}.

Adım Adım Çözüm

1
Identify the formula for magnetic force on a moving charge
F=qvBsin⁡θF = qvB\sin\theta
A charged particle moving at an angle through a magnetic field experiences a force proportional to the velocity component perpendicular to the field.
2
Substitute the given physical values into the equation
3.2×10−13=(1.6×10−19)×(4.0×106)×B×sin⁡(30∘)3.2 \times 10^{-13} = (1.6 \times 10^{-19}) \times (4.0 \times 10^6) \times B \times \sin(30^\circ)
Values provided: charge q=1.6×10−19 Cq = 1.6 \times 10^{-19}\text{ C}, velocity v=4.0×106 m s−1v = 4.0 \times 10^6\text{ m s}^{-1}, force F=3.2×10−13 NF = 3.2 \times 10^{-13}\text{ N}, and angle θ=30∘\theta = 30^\circ.
3
Evaluate sin⁡(30∘)\sin(30^\circ) and simplify the numerical product
3.2×10−13=(1.6×10−19)×(4.0×106)×0.5×B=3.2×10−13×B3.2 \times 10^{-13} = (1.6 \times 10^{-19}) \times (4.0 \times 10^6) \times 0.5 \times B = 3.2 \times 10^{-13} \times B
Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, multiplying 1.6×10−19×4.0×106×0.51.6 \times 10^{-19} \times 4.0 \times 10^6 \times 0.5 yields 3.2×10−133.2 \times 10^{-13}.
4
Solve for the magnetic flux density BB
B=3.2×10−133.2×10−13=1.0 TB = \frac{3.2 \times 10^{-13}}{3.2 \times 10^{-13}} = 1.0\text{ T}
Dividing both sides of the equation by 3.2×10−13 N C−1 m−1 s3.2 \times 10^{-13}\text{ N C}^{-1}\text{ m}^{-1}\text{ s} yields the magnetic flux density.

Anahtar Kavram

Magnetic Force on a Moving Charge
Bu soruyu puanla

Konular

Magnetic Force and Electromagnetism

Bu konudan daha fazla soru