An electron that has a velocity with component 2.7 × 10^6 m/s in the x direction and 4.3 × 10^6 m/s in the y direction moves through a uniform magnetic field with x component 0.023 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat the calculation for a proton having the same velocity.
Solution
The problem presented involves finding the magnetic force on an electron moving through a magnetic field and then repeating the calculation for a proton with the same velocity.
We use the formula for the magnetic force on a charged particle:
F=q(v×B)
Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle (in vector form),
B is the magnetic field (also in vector form),
× represents the cross product.
Given dataThe problem presented involves finding the magnetic force on an electron moving through a magnetic field and then repeating the calculation for a proton with the same velocity.
We use the formula for the magnetic force on a charged particle:
F=q(v×B)
Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle (in vector form),
B is the magnetic field (also in vector form),
× represents the cross product.
Given
Velocity of the electron: v=(2.7×106m/s)i^+(4.3×106m/s)j^
Magnetic field: B=(0.023T)i^+(−0.12T)j^
Charge of an electron: qe=−1.6×10−19C
Charge of a proton: qp=+1.6×10−19C
Part (a): Magnetic force on the electron
We will first calculate the cross product v×B and then multiply by the charge qe.
Cross productv×B:
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2.7 \times 10^6 & 4.3 \times 10^6 & 0 \\
0.023 & -0.12 & 0
\end{vmatrix}$$
This yields:
$$\mathbf{v} \times \mathbf{B} = (0 \, \hat{i} + 0 \, \hat{j} + (-2.7 \times 10^6 \times -0.12 - 4.3 \times 10^6 \times 0.023) \hat{k})$$
Simplifying:
$$\mathbf{v} \times \mathbf{B} = (0 + 0 + (3.24 \times 10^5 + (-9.89 \times 10^4)) \hat{k})$$
$$\mathbf{v} \times \mathbf{B} = (2.251 \times 10^5) \hat{k} \, \text{m/s·T}$$
2. **Magnetic force** on the electron:
$$\mathbf{F}_e = q_e (\mathbf{v} \times \mathbf{B}) = -1.6 \times 10^{-19} \times (2.251 \times 10^5) \hat{k}$$
$$\mathbf{F}_e = -3.6016 \times 10^{-14} \, \text{N} \, \hat{k}$$
The magnitude of the force is:
$$|\mathbf{F}_e| = 3.6016 \times 10^{-14} \, \text{N}$$
### Part (b): Magnetic force on a proton
Since the proton has the same velocity and the same magnetic field, the only difference is the sign of the charge. Therefore, the magnetic force on the proton will be the same magnitude but with opposite direction:
$$\mathbf{F}_p = +3.6016 \times 10^{-14} \, \text{N} \, \hat{k}$$
Thus, the magnitude of the force on the proton is also:
$$|\mathbf{F}_p| = 3.6016 \times 10^{-14} \, \text{N}$$
Would you like further details or have any questions? Here are some related questions to expand on this problem:
1. How does the direction of the magnetic force depend on the sign of the charge?
2. What is the physical meaning of the cross product in this context?
3. How would the result change if the velocity had a component in the z-direction?
4. How can we determine the direction of the magnetic force using the right-hand rule?
5. What happens to the path of the electron and proton in the magnetic field?
**Tip:** Remember that the force on a particle in a magnetic field depends on both the velocity and the direction of the magnetic field, as well as the charge of the particle.