Question

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

Answer 1

Correct Answer: 5

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 \([5, 5]\).

Answer 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:

\[ B_0 = 5T \]