An electron moves through a uniform magnetic field \( \vec{B} = B_0 \hat{i} + 2B_0 \hat{j} \, \text{T} \). At a particular instant of time, the velocity of the electron is \( \vec{u} = 3\hat{i} + 5\hat{j} \, \text{m/s} \). If the magnetic force acting on the electron is \( \vec{F} = 5e k \, \text{N} \), where \( e \) is the charge of the electron, then the value of \( B_0 \) is _____ T
Correct Answer: 5
Approach Solution - 1
The magnetic force on a charged particle moving through a magnetic field is given by the formula:
\(|\vec{F}| = q|\vec{u} \times \vec{B}|\)
Given:\( \vec{F} = 5e \hat{k} \, \text{N} \), \( q = -e \) (charge of electron), \( \vec{u} = 3\hat{i} + 5\hat{j} \, \text{m/s} \), and \( \vec{B} = B_0 \hat{i} + 2B_0 \hat{j} \, \text{T} \).
Calculate cross product \(\vec{u} \times \vec{B}\):
\(|\vec{u} \times \vec{B}| = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 0 \\ B_0 & 2B_0 & 0 \end{array}\right|\)
The determinant results in \(\hat{0}\hat{i} - \hat{0}\hat{j} + (3 \cdot 2B_0 - 5 \cdot B_0) \hat{k}\):
\(|\vec{u} \times \vec{B}| = (6B_0 - 5B_0)\hat{k} = B_0\hat{k}\)
Thus, \(|\vec{F}| = e|\vec{u} \times \vec{B}| = eB_0\)
Since \(|\vec{F}| = 5e\), equating gives:
\(eB_0 = 5e\)
Solving for \(B_0\), we find:
\(B_0 = 5\)
The solution \(B_0 = 5 \, \text{T}\) is within the given range \([5, 5]\).
Approach Solution -2
The magnetic force is given by:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
Substituting the values:
\[ 5e \hat{k} = e \left( 3\hat{i} + 5\hat{j} \right) \times \left( B_0 \hat{i} + 2B_0 \hat{j} \right) \]
Calculating the cross product:
\[ 5e \hat{k} = e \left( 6B_0 \hat{k} - 5B_0 \hat{k} \right) \]
Simplifying:
\[ 5e \hat{k} = e B_0 \hat{k} \]
Therefore:
\[ B_0 = 5T \]
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