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Question: An electric charge + q moves with velocity → v = 3 ˆ i + 4 ˆ j + ˆ k , in an electromagnetic fiel...

An electric charge + q moves with velocity → v = 3 ˆ i + 4 ˆ j + ˆ k , in an electromagnetic field given by → E = 3 ˆ i + ˆ j + 2 ˆ k , → B = ˆ i + ˆ j − 3 ˆ k . The y-component of the force experienced by +q is

Answer

11q

Explanation

Solution

The force experienced by a charge qq moving with velocity v⃗\vec{v} in an electromagnetic field (E⃗\vec{E} and B⃗\vec{B}) is given by the Lorentz force law:

F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})

Given:

Charge = +q+q

Velocity v⃗=3i^+4j^+k^\vec{v} = 3 \hat{i} + 4 \hat{j} + \hat{k}

Electric field E⃗=3i^+j^+2k^\vec{E} = 3 \hat{i} + \hat{j} + 2 \hat{k}

Magnetic field B⃗=i^+j^−3k^\vec{B} = \hat{i} + \hat{j} - 3 \hat{k}

First, calculate the electric force component F⃗E\vec{F}_E:

F⃗E=qE⃗=q(3i^+j^+2k^)=3qi^+qj^+2qk^\vec{F}_E = q\vec{E} = q(3 \hat{i} + \hat{j} + 2 \hat{k}) = 3q \hat{i} + q \hat{j} + 2q \hat{k}

Next, calculate the cross product v⃗×B⃗\vec{v} \times \vec{B}:

v⃗×B⃗=∣i^j^k^34111−3∣\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 1 \\ 1 & 1 & -3 \end{vmatrix} =i^((4)(−3)−(1)(1))−j^((3)(−3)−(1)(1))+k^((3)(1)−(4)(1))= \hat{i}((4)(-3) - (1)(1)) - \hat{j}((3)(-3) - (1)(1)) + \hat{k}((3)(1) - (4)(1)) =i^(−12−1)−j^(−9−1)+k^(3−4)= \hat{i}(-12 - 1) - \hat{j}(-9 - 1) + \hat{k}(3 - 4) =−13i^+10j^−k^= -13 \hat{i} + 10 \hat{j} - \hat{k}

Now, calculate the magnetic force component F⃗B\vec{F}_B:

F⃗B=q(v⃗×B⃗)=q(−13i^+10j^−k^)=−13qi^+10qj^−qk^\vec{F}_B = q(\vec{v} \times \vec{B}) = q(-13 \hat{i} + 10 \hat{j} - \hat{k}) = -13q \hat{i} + 10q \hat{j} - q \hat{k}

Finally, add the electric and magnetic force components to find the total force F⃗\vec{F}:

F⃗=F⃗E+F⃗B\vec{F} = \vec{F}_E + \vec{F}_B F⃗=(3qi^+qj^+2qk^)+(−13qi^+10qj^−qk^)\vec{F} = (3q \hat{i} + q \hat{j} + 2q \hat{k}) + (-13q \hat{i} + 10q \hat{j} - q \hat{k}) F⃗=(3q−13q)i^+(q+10q)j^+(2q−q)k^\vec{F} = (3q - 13q) \hat{i} + (q + 10q) \hat{j} + (2q - q) \hat{k} F⃗=−10qi^+11qj^+qk^\vec{F} = -10q \hat{i} + 11q \hat{j} + q \hat{k}

The y-component of the force is the coefficient of j^\hat{j}, which is 11q11q.

In summary:

The total force on a charge in an electromagnetic field is the sum of the electric force (qE⃗q\vec{E}) and the magnetic force (q(v⃗×B⃗)q(\vec{v} \times \vec{B})).

  1. Calculate the electric force vector F⃗E\vec{F}_E.

  2. Calculate the cross product v⃗×B⃗\vec{v} \times \vec{B}.

  3. Calculate the magnetic force vector F⃗B\vec{F}_B.

  4. Add F⃗E\vec{F}_E and F⃗B\vec{F}_B to get the total force vector F⃗\vec{F}.

  5. Identify the y-component from the total force vector.