An electron moves through a uniform magnetic field given by B = Bx i + (3.0Bx) j. At a particular instant, the electron has velocity v = (2.0i + 4.0) j m/s and the magnetic force acting on it is (6.4 x 10-19N) k. Find Bx
The Correct Answer and Explanation is :
To solve for ( B_x ), we use the formula for the magnetic force on a charged particle moving through a magnetic field:
[
\vec{F} = q \, (\vec{v} \times \vec{B})
]
Step 1: Given data
- Charge of the electron: ( q = -1.6 \times 10^{-19} \, \text{C} )
- Velocity: ( \vec{v} = 2.0 \, \hat{i} + 4.0 \, \hat{j} \, \text{m/s} )
- Magnetic field: ( \vec{B} = B_x \, \hat{i} + 3.0 B_x \, \hat{j} )
- Magnetic force: ( \vec{F} = 6.4 \times 10^{-19} \, \hat{k} \, \text{N} )
Step 2: Cross product of ( \vec{v} ) and ( \vec{B} )
Using the determinant form for the cross product:
[
\vec{v} \times \vec{B} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \
2.0 & 4.0 & 0 \
B_x & 3.0B_x & 0
\end{vmatrix}
]
Expanding the determinant:
[
\vec{v} \times \vec{B} = \hat{i} \left(4.0 \cdot 0 – 0 \cdot 3.0B_x \right)
- \hat{j} \left(2.0 \cdot 0 – 0 \cdot B_x \right)
- \hat{k} \left(2.0 \cdot 3.0B_x – 4.0 \cdot B_x \right)
]
[
\vec{v} \times \vec{B} = \hat{k} \left(6.0B_x – 4.0B_x\right)
]
[
\vec{v} \times \vec{B} = 2.0B_x \, \hat{k}
]
Step 3: Relating ( \vec{F} ) to ( \vec{v} \times \vec{B} )
The magnetic force is:
[
\vec{F} = q \, (\vec{v} \times \vec{B})
]
Substitute ( q = -1.6 \times 10^{-19} \, \text{C} ) and ( \vec{v} \times \vec{B} = 2.0B_x \, \hat{k} ):
[
\vec{F} = (-1.6 \times 10^{-19})(2.0B_x) \, \hat{k}
]
[
\vec{F} = -3.2 \times 10^{-19}B_x \, \hat{k}
]
Step 4: Equate to given force
The given force is ( \vec{F} = 6.4 \times 10^{-19} \, \hat{k} ). Comparing magnitudes:
[
|-3.2 \times 10^{-19}B_x| = 6.4 \times 10^{-19}
]
[
3.2 \times 10^{-19}B_x = 6.4 \times 10^{-19}
]
Solve for ( B_x ):
[
B_x = \frac{6.4 \times 10^{-19}}{3.2 \times 10^{-19}} = 2.0
]
Final Answer:
[
B_x = 2.0 \, \text{T}
]
Explanation:
The problem involves finding the component of the magnetic field (( B_x )) that results in the given magnetic force on the electron. By taking the cross product of velocity and magnetic field, we derived the magnetic force and used the given force magnitude to solve for ( B_x ). The negative electron charge reversed the direction, and the magnitude calculation yielded ( B_x = 2.0 \, \text{T} ).