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Question: A electron experiences a force ( 4.0 ˆ i + 3.0 ˆ j ) × 10 − 13 N in a uniform magnetic field when ...

A electron experiences a force ( 4.0 ˆ i + 3.0 ˆ j ) × 10 − 13 N in a uniform magnetic field when its velocity is 2.5 h a k × 10 7 m s − 1 When the velocity is redirected and becomes ( 1.5 ˆ i − 2.0 ˆ j ) × 10 7 × 10 − 7 the magnetic force of the electron is zero. The magnetic field vectro → B is .

Answer

The magnetic field vector B⃗\vec{B} is (−0.075i^+0.100j^) T(-0.075 \hat{i} + 0.100 \hat{j}) \text{ T}.

Explanation

Solution

The problem involves calculating the magnetic field vector given the magnetic force on an electron at two different velocities. The key principle is the Lorentz force law: F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}).

  1. Deduce the direction of B⃗\vec{B}:

    The problem states that when the electron's velocity is v⃗2=(1.5i^−2.0j^)×107 m/s\vec{v}_2 = (1.5 \hat{i} - 2.0 \hat{j}) \times 10^7 \text{ m/s} (assuming the typo "×107×10−7\times 10^7 \times 10^{-7}" means ×107\times 10^7), the magnetic force F⃗2=0\vec{F}_2 = 0.

    For the magnetic force to be zero on a moving charge (q≠0,v⃗≠0q \neq 0, \vec{v} \neq 0), the velocity vector must be parallel or anti-parallel to the magnetic field vector (v⃗×B⃗=0\vec{v} \times \vec{B} = 0).

    Therefore, B⃗\vec{B} must be parallel to v⃗2\vec{v}_2. This implies that B⃗\vec{B} has no zz-component and its xx and yy components are proportional to those of v⃗2\vec{v}_2.

    Let B⃗=k(1.5i^−2.0j^)\vec{B} = k (1.5 \hat{i} - 2.0 \hat{j}) for some scalar constant kk.

  2. Use the first scenario to find the value of kk:

    In the first scenario, the electron experiences a force F⃗1=(4.0i^+3.0j^)×10−13 N\vec{F}_1 = (4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-13} \text{ N} when its velocity is v⃗1=2.5×107k^ m/s\vec{v}_1 = 2.5 \times 10^7 \hat{k} \text{ m/s} (assuming "h a k" means k^\hat{k}).

    The charge of an electron is q=−e=−1.6×10−19 Cq = -e = -1.6 \times 10^{-19} \text{ C}.

    Using the Lorentz force law: F⃗1=q(v⃗1×B⃗)\vec{F}_1 = q (\vec{v}_1 \times \vec{B})

    Substitute the known values:

    (4.0i^+3.0j^)×10−13=(−e)[(2.5×107k^)×k(1.5i^−2.0j^)](4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-13} = (-e) [(2.5 \times 10^7 \hat{k}) \times k(1.5 \hat{i} - 2.0 \hat{j})]

    First, calculate the cross product v⃗1×B⃗\vec{v}_1 \times \vec{B}:

    v⃗1×B⃗=(2.5×107k^)×k(1.5i^−2.0j^)\vec{v}_1 \times \vec{B} = (2.5 \times 10^7 \hat{k}) \times k(1.5 \hat{i} - 2.0 \hat{j})

    =(2.5×107)k[1.5(k^×i^)−2.0(k^×j^)]= (2.5 \times 10^7) k [1.5 (\hat{k} \times \hat{i}) - 2.0 (\hat{k} \times \hat{j})]

    =(2.5×107)k[1.5j^−2.0(−i^)]= (2.5 \times 10^7) k [1.5 \hat{j} - 2.0 (-\hat{i})]

    =(2.5×107)k[2.0i^+1.5j^]= (2.5 \times 10^7) k [2.0 \hat{i} + 1.5 \hat{j}]

    Now, substitute this back into the force equation:

    (4.0i^+3.0j^)×10−13=(−e)(2.5×107)k[2.0i^+1.5j^](4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-13} = (-e) (2.5 \times 10^7) k [2.0 \hat{i} + 1.5 \hat{j}]

    (4.0i^+3.0j^)×10−13=[−e(2.5×107)(2.0)k]i^+[−e(2.5×107)(1.5)k]j^(4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-13} = [-e (2.5 \times 10^7) (2.0) k] \hat{i} + [-e (2.5 \times 10^7) (1.5) k] \hat{j}

    (4.0i^+3.0j^)×10−13=(−5.0×107ek)i^+(−3.75×107ek)j^(4.0 \hat{i} + 3.0 \hat{j}) \times 10^{-13} = (-5.0 \times 10^7 e k) \hat{i} + (-3.75 \times 10^7 e k) \hat{j}

  3. Solve for kk:

    Equate the coefficients of i^\hat{i} and j^\hat{j}:

    For i^\hat{i}: 4.0×10−13=−5.0×107ek4.0 \times 10^{-13} = -5.0 \times 10^7 e k

    ek=4.0×10−13−5.0×107=4.0×10−20−5.0=−0.8×10−20e k = \frac{4.0 \times 10^{-13}}{-5.0 \times 10^7} = \frac{4.0 \times 10^{-20}}{-5.0} = -0.8 \times 10^{-20}

    For j^\hat{j}: 3.0×10−13=−3.75×107ek3.0 \times 10^{-13} = -3.75 \times 10^7 e k

    ek=3.0×10−13−3.75×107=3.0×10−20−3.75=−0.8×10−20e k = \frac{3.0 \times 10^{-13}}{-3.75 \times 10^7} = \frac{3.0 \times 10^{-20}}{-3.75} = -0.8 \times 10^{-20}

    The values are consistent.

    Now, calculate kk using e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}:

    k=−0.8×10−201.6×10−19=−0.81.6×10−20+19=−0.5×10−1=−0.05k = \frac{-0.8 \times 10^{-20}}{1.6 \times 10^{-19}} = -\frac{0.8}{1.6} \times 10^{-20+19} = -0.5 \times 10^{-1} = -0.05

  4. Determine the magnetic field vector B⃗\vec{B}:

    Substitute the value of kk back into the expression for B⃗\vec{B}:

    B⃗=k(1.5i^−2.0j^)\vec{B} = k (1.5 \hat{i} - 2.0 \hat{j})

    B⃗=−0.05(1.5i^−2.0j^)\vec{B} = -0.05 (1.5 \hat{i} - 2.0 \hat{j})

    B⃗=(−0.05×1.5i^+0.05×2.0j^)\vec{B} = (-0.05 \times 1.5 \hat{i} + 0.05 \times 2.0 \hat{j})

    B⃗=(−0.075i^+0.100j^) T\vec{B} = (-0.075 \hat{i} + 0.100 \hat{j}) \text{ T}

Explanation of the solution:

  1. The condition F⃗2=0\vec{F}_2 = 0 implies that B⃗\vec{B} is parallel to v⃗2\vec{v}_2. This fixes the direction of B⃗\vec{B} as proportional to (1.5i^−2.0j^)(1.5 \hat{i} - 2.0 \hat{j}).
  2. The Lorentz force equation F⃗1=q(v⃗1×B⃗)\vec{F}_1 = q(\vec{v}_1 \times \vec{B}) is then used.
  3. Substitute the known values for F⃗1\vec{F}_1, v⃗1\vec{v}_1, qq, and the proportional form of B⃗\vec{B} into the equation.
  4. Perform the vector cross product and equate the components of the resulting force vector with the given F⃗1\vec{F}_1.
  5. Solve the resulting scalar equations to find the proportionality constant, kk.
  6. Substitute kk back into the expression for B⃗\vec{B} to get the final magnetic field vector.