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Question: An electron traveling west to east enters a chamber having a uniform electrostatic field in north to...

An electron traveling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Answer

Vertically downwards

Explanation

Solution

The problem describes an electron moving in perpendicular electric and magnetic fields, and asks for the direction of the magnetic field required to prevent its deflection. For the electron to continue in a straight line, the net force on it must be zero. This means the electric force must be balanced by an equal and opposite magnetic force.

Let's define a coordinate system:

  • East: +x direction
  • North: +y direction
  • Up: +z direction

From this, West is -x, South is -y, and Down is -z.

  1. Velocity of the electron (v⃗\vec{v}): The electron travels West to East, so its velocity is in the +x direction. v⃗=vi^\vec{v} = v \hat{i}

  2. Electrostatic field (E⃗\vec{E}): The field is in the North to South direction, so it's in the -y direction. E⃗=−Ej^\vec{E} = -E \hat{j}

  3. Electric Force (F⃗E\vec{F}_E): The force on a charged particle in an electric field is F⃗E=qE⃗\vec{F}_E = q\vec{E}. Since the electron has a negative charge (q=−eq = -e): F⃗E=(−e)(−Ej^)=eEj^\vec{F}_E = (-e)(-E \hat{j}) = eE \hat{j} This means the electric force on the electron is in the +y direction (South to North).

  4. Required Magnetic Force (F⃗B\vec{F}_B): To prevent deflection, the magnetic force must be equal in magnitude and opposite in direction to the electric force. F⃗B=−F⃗E=−eEj^\vec{F}_B = -\vec{F}_E = -eE \hat{j} So, the magnetic force must be in the -y direction (North to South).

  5. Direction of Magnetic Field (B⃗\vec{B}): The magnetic force on a charged particle is given by the Lorentz force law: F⃗B=q(v⃗×B⃗)\vec{F}_B = q(\vec{v} \times \vec{B}). Substitute the known values and the required magnetic force: −eEj^=(−e)(vi^×B⃗)-eE \hat{j} = (-e) (v \hat{i} \times \vec{B}) Dividing by −e-e (assuming e≠0e \neq 0): Ej^=v(i^×B⃗)E \hat{j} = v (\hat{i} \times \vec{B})

    We need to find a direction for B⃗\vec{B} such that when i^\hat{i} (East) is crossed with B⃗\vec{B}, the result is in the j^\hat{j} (North) direction. Using the right-hand rule for the cross product A⃗×B⃗\vec{A} \times \vec{B}: point your fingers in the direction of A⃗\vec{A} (East), and curl them towards the direction of B⃗\vec{B}. Your thumb will point in the direction of A⃗×B⃗\vec{A} \times \vec{B}. Here, A⃗=i^\vec{A} = \hat{i} (East) and the result A⃗×B⃗\vec{A} \times \vec{B} must be j^\hat{j} (North). If you point your fingers East and want your thumb to point North, you must curl your fingers downwards. Therefore, the magnetic field B⃗\vec{B} must be directed vertically downwards.

    Let's verify this using vector components: Let B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}. i^×B⃗=i^×(Bxi^+Byj^+Bzk^)\hat{i} \times \vec{B} = \hat{i} \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) =Bx(i^×i^)+By(i^×j^)+Bz(i^×k^)= B_x (\hat{i} \times \hat{i}) + B_y (\hat{i} \times \hat{j}) + B_z (\hat{i} \times \hat{k}) =0+Byk^+Bz(−j^)= 0 + B_y \hat{k} + B_z (-\hat{j}) =Byk^−Bzj^= B_y \hat{k} - B_z \hat{j}

    Comparing this with Ej^=v(i^×B⃗)E \hat{j} = v (\hat{i} \times \vec{B}): Ej^=v(Byk^−Bzj^)E \hat{j} = v (B_y \hat{k} - B_z \hat{j}) Comparing the components: For k^\hat{k}: 0=vBy  ⟹  By=00 = v B_y \implies B_y = 0 For j^\hat{j}: E=−vBz  ⟹  Bz=−E/vE = -v B_z \implies B_z = -E/v

    This confirms that the magnetic field must be in the −k^-\hat{k} direction, which is vertically downwards.

Explanation of the solution:

  1. Determine the direction of the electric force (F⃗E=qE⃗\vec{F}_E = q\vec{E}) on the electron. Since the electron is negatively charged and the electric field is North to South, the electric force is South to North.
  2. For no deflection, the magnetic force (F⃗B\vec{F}_B) must be equal and opposite to the electric force. Thus, the magnetic force must be North to South.
  3. Use the Lorentz force formula (F⃗B=q(v⃗×B⃗)\vec{F}_B = q(\vec{v} \times \vec{B})) and the right-hand rule. For a negative charge, the direction of F⃗B\vec{F}_B is opposite to the direction of v⃗×B⃗\vec{v} \times \vec{B}.
  4. The electron's velocity (v⃗\vec{v}) is East. The desired magnetic force (F⃗B\vec{F}_B) is North to South. Therefore, v⃗×B⃗\vec{v} \times \vec{B} must be South to North.
  5. Applying the right-hand rule (fingers along East, thumb along South to North), the magnetic field (B⃗\vec{B}) must point vertically downwards.