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CHAPTER FIVE: MAGNETIC FIELD Calculate the magnetic field strength exerting a force of 9x10-19N on a charge of 1.6x10-19C travelling at a speed of 3x106m/s perpendicular to the field?

Physics

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1. Calculate the magnetic field strength exerting a force of 9×10−19 N9 \times 10^{-19} \text{ N} on a charge of 1.6×10−19 C1.6 \times 10^{-19} \text{ C} travelling at a speed of 3×106 m/s3 \times 10^6 \text{ m/s} perpendicular to the field.

The magnetic force (FF) on a charge (qq) moving with velocity (vv) in a magnetic field (BB) is given by F=qvBsin⁡θF = qvB\sin\theta. Since the motion is perpendicular, θ=90∘\theta = 90^\circ, so sin⁡θ=1\sin\theta = 1.

  1. Write down the formula and given values.

    F=qvBF = qvB F=9×10−19 NF = 9 \times 10^{-19} \text{ N} q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C} v=3×106 m/sv = 3 \times 10^6 \text{ m/s}
  2. Rearrange the formula to solve for BB and substitute the values.

    B=FqvB = \frac{F}{qv} B=9×10−19 N(1.6×10−19 C)(3×106 m/s)B = \frac{9 \times 10^{-19} \text{ N}}{(1.6 \times 10^{-19} \text{ C})(3 \times 10^6 \text{ m/s})}
  3. Calculate the magnetic field strength.

    B=9×10−194.8×10−13 TB = \frac{9 \times 10^{-19}}{4.8 \times 10^{-13}} \text{ T} B=1.875×10−6 TB = 1.875 \times 10^{-6} \text{ T}

    The magnetic field strength is 1.875×10−6 T\boxed{\mathbf{1.875 \times 10^{-6} \text{ T}}}.

    2. Calculate the magnitude of the force due to a magnetic field of 0.40 T0.40 \text{ T} on an electron moving perpendicular to the field with a speed of 4.00×106 m/s4.00 \times 10^6 \text{ m/s}. (e=1.60×10−19 Ce = 1.60 \times 10^{-19} \text{ C})

    The magnetic force (FF) on a charge (qq) moving with velocity (vv) in a magnetic field (BB) is given by F=qvBsin⁡θF = qvB\sin\theta. Since the motion is perpendicular, θ=90∘\theta = 90^\circ, so sin⁡θ=1\sin\theta = 1.

  4. Write down the formula and given values.

    F=qvBF = qvB q=e=1.60×10−19 Cq = e = 1.60 \times 10^{-19} \text{ C} v=4.00×106 m/sv = 4.00 \times 10^6 \text{ m/s} B=0.40 TB = 0.40 \text{ T}
  5. Substitute the values into the formula.

    F=(1.60×10−19 C)(4.00×106 m/s)(0.40 T)F = (1.60 \times 10^{-19} \text{ C})(4.00 \times 10^6 \text{ m/s})(0.40 \text{ T})
  6. Calculate the magnitude of the force.

    F=2.56×10−13 NF = 2.56 \times 10^{-13} \text{ N}

    The magnitude of the force is 2.56×10−13 N\boxed{\mathbf{2.56 \times 10^{-13} \text{ N}}}.

    3. With a defined diagram explain magnetic field for both internal and external.

    A magnetic field is a region around a magnet or a current-carrying conductor where magnetic forces can be detected. It is represented by magnetic field lines which originate from the North pole and end at the South pole outside the magnet, and run from South to North inside the magnet, forming continuous closed loops.

    • External Magnetic Field: This is the field observed outside a magnet or a current-carrying object. For a bar magnet, field lines emerge from the North pole and curve around to enter the South pole. For a straight current-carrying wire, the field lines form concentric circles around the wire.
    • Internal Magnetic Field: This is the field present within the material of a magnet or inside a current-carrying coil (like a solenoid). Inside a bar magnet, the field lines run from the South pole to the North pole, completing the loop. Inside a solenoid, the field lines are nearly uniform and parallel to the axis of the coil.
          N -------------------- S
          |                    |
          |  <---------------- |  (Internal field lines S to N)
          |                    |
          |  ----------------> |  (External field lines N to S)
          |                    |
          ----------------------
    

    (Imagine the lines curving outside from N to S and straight inside from S to N)

    4. A long straight wire carries a current of 35 Amp35 \text{ Amp}. What is the magnitude of the field B at a point 20 cm20 \text{ cm} from the wire?

    The magnetic field (BB) at a distance (rr) from a long straight wire carrying current (II) is given by B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}.

  7. Convert the distance to meters and write down the given values.

    r=20 cm=0.20 mr = 20 \text{ cm} = 0.20 \text{ m} I=35 AI = 35 \text{ A} μ0=4π×10−7T⋅mA\mu_0 = 4\pi \times 10^{-7} \frac{\text{T} \cdot \text{m}}{\text{A}}
  8. Substitute the values into the formula.

    B=(4π×10−7T⋅mA)(35 A)2π(0.20 m)B = \frac{(4\pi \times 10^{-7} \frac{\text{T} \cdot \text{m}}{\text{A}}) (35 \text{ A})}{2\pi (0.20 \text{ m})}
  9. Calculate the magnetic field magnitude.

    B=140π×10−70.40π TB = \frac{140\pi \times 10^{-7}}{0.40\pi} \text{ T} B=140×10−70.40 TB = \frac{140 \times 10^{-7}}{0.40} \text{ T} B=3.5×10−5 TB = 3.5 \times 10^{-5} \text{ T}

    The magnitude of the field B is 3.5×10−5 T\boxed{\mathbf{3.5 \times 10^{-5} \text{ T}}}.

    5. A horizontal overhead power line carries a current of 90 Amp90 \text{ Amp} in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m1.5 \text{ m} below the line?

    The magnitude of the magnetic field (BB) is given by B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}. The direction is found using the Right-Hand Rule.

  10. Write down the given values.

    I=90 AI = 90 \text{ A} r=1.5 mr = 1.5 \text{ m} μ0=4π×10−7T⋅mA\mu_0 = 4\pi \times 10^{-7} \frac{\text{T} \cdot \text{m}}{\text{A}}
  11. Substitute the values into the formula to calculate the magnitude.

    B=(4π×10−7T⋅mA)(90 A)2π(1.5 m)B = \frac{(4\pi \times 10^{-7} \frac{\text{T} \cdot \text{m}}{\text{A}}) (90 \text{ A})}{2\pi (1.5 \text{ m})} B=360π×10−73π TB = \frac{360\pi \times 10^{-7}}{3\pi} \text{ T} B=120×10−7 TB = 120 \times 10^{-7} \text{ T} B=1.2×10−5 TB = 1.2 \times 10^{-5} \text{ T}
  12. Determine the direction using the Right-Hand Rule.

    Point your right thumb in the direction of the current (East to West). Curl your fingers. Below the wire, your fingers will point towards the North.
    The magnitude of the magnetic field is 1.2×10−5 T\boxed{\mathbf{1.2 \times 10^{-5} \text{ T}}} and its direction is North.

    6. A square coil of side 10 cm10 \text{ cm} consists of 20 turns and carries a current of 12 A12 \text{ A}. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30∘30^\circ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T0.80 \text{ T}. What is the magnitude of torque experienced by the coil?

    The torque (τ\tau) on a current loop in a magnetic field is given by τ=NIABsin⁡θ\tau = NIAB\sin\theta.

  13. Convert the side length to meters and calculate the area.

    s=10 cm=0.10 ms = 10 \text{ cm} = 0.10 \text{ m} A=s2=(0.10 m)2=0.01 m2A = s^2 = (0.10 \text{ m})^2 = 0.01 \text{ m}^2
  14. Write down the given values.

    N=20 turnsN = 20 \text{ turns} I=12 AI = 12 \text{ A} B=0.80 TB = 0.80 \text{ T} θ=30∘\theta = 30^\circ
  15. Substitute the values into the torque formula.

    τ=(20)(12 A)(0.01 m2)(0.80 T)sin⁡(30∘)\tau = (20)(12 \text{ A})(0.01 \text{ m}^2)(0.80 \text{ T})\sin(30^\circ) τ=(20)(12)(0.01)(0.80)(0.5) N⋅m\tau = (20)(12)(0.01)(0.80)(0.5) \text{ N} \cdot \text{m}
  16. Calculate the magnitude of the torque.

    τ=0.96 N⋅m\tau = 0.96 \text{ N} \cdot \text{m}

    The magnitude of torque experienced by the coil is 0.96 N⋅m\boxed{\mathbf{0.96 \text{ N} \cdot \text{m}}}.

    7. Two concentric circular coils X and Y radii 16 cm16 \text{ cm} and 10 cm10 \text{ cm}, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A16 \text{ A}; coil Y has 25 turns and carries a current of 18 A18 \text{ A}. The sense of the current...

    The question is cut off, but it most likely asks for the net magnetic field at the center of the coils. I will assume the currents in both coils are in the same direction (e.g., both clockwise or both counter-clockwise) so their magnetic fields add up. If they were in opposite directions, the fields would subtract.

    The magnetic field (BB) at the center of a circular coil is given by B=μ0NI2rB = \frac{\mu_0 N I}{2r}.

  17. Calculate the magnetic field due to Coil X.

    NX=20N_X = 20 rX=16 cm=0.16 mr_X = 16 \text{ cm} = 0.16 \text{ m} IX=16 AI_X = 16 \text{ A} BX=(4π×10−7T⋅mA)(20)(16 A)2(0.16 m)B_X = \frac{(4\pi \times 10^{-7} \frac{\text{T} \cdot \text{m}}{\text{A}}) (20) (16 \text{ A})}{2 (0.16 \text{ m})} BX=1280π×10−70.32 TB_X = \frac{1280\pi \times 10^{-7}}{0.32} \text{ T} BX=4000π×10−7 T=4π×10−4 TB_X = 4000\pi \times 10^{-7} \text{ T} = 4\pi \times 10^{-4} \text{ T}
  18. Calculate the magnetic field due to Coil Y.

    NY=25N_Y = 25 rY=10 cm=0.10 mr_Y = 10 \text{ cm} = 0.10 \text{ m} IY=18 AI_Y = 18 \text{ A} BY=(4π×10−7T⋅mA)(25)(18 A)2(0.10 m)B_Y = \frac{(4\pi \times 10^{-7} \frac{\text{T} \cdot \text{m}}{\text{A}}) (25) (18 \text{ A})}{2 (0.10 \text{ m})} BY=1800π×10−70.20 TB_Y = \frac{1800\pi \times 10^{-7}}{0.20} \text{ T} BY=9000π×10−7 T=9π×10−4 TB_Y = 9000\pi \times 10^{-7} \text{ T} = 9\pi \times 10^{-4} \text{ T}
  19. Calculate the net magnetic field at the center (assuming currents are in the same direction).

    Bnet=BX+BYB_{net} = B_X + B_Y Bnet=(4π×10−4 T)+(9π×10−4 T)B_{net} = (4\pi \times 10^{-4} \text{ T}) + (9\pi \times 10^{-4} \text{ T}) Bnet=13π×10−4 TB_{net} = 13\pi \times 10^{-4} \text{ T} Bnet≈13×3.14159×10−4 TB_{net} \approx 13 \times 3.14159 \times 10^{-4} \text{ T} Bnet≈4.08×10−3 TB_{net} \approx 4.08 \times 10^{-3} \text{ T}

    The net magnetic field at the center (assuming currents are in the same direction) is 4.08×10−3 T\boxed{\mathbf{4.08 \times 10^{-3} \text{ T}}}.

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