Question

A particle of charge \(1.0\times10^{-16}\ \text{C}\) moves through a uniform magnetic field \[ \vec{B}=B_0(\hat{i}+4\hat{j})\ \text{T}. \] The particle velocity at some instant is \[ \vec{V}=(2\hat{i}+4\hat{j})\ \text{m s}^{-1} \] and the magnetic force acting on it is \[ 3\times10^{-16}\hat{k}\ \text{N}. \] The magnitude of \(B_0\) is

• A. \(1.0\ \text{T}\)
• B. \(2.5\ \text{T}\)
• C. \(0.5\ \text{T}\)
• D. \(0.75\ \text{T}\)

Hint:

For magnetic force problems, always use \[ \vec{F}=q(\vec{V}\times\vec{B}). \] When vectors are given in component form, evaluate the cross product using a determinant. The direction is obtained from the right-hand rule.

Answer 1

The Correct Option is D

Step 1: Use the magnetic force formula.
The magnetic force on a charged particle is given by \[ \vec{F}=q(\vec{V}\times\vec{B}) \] Given, \[ q=1.0\times10^{-16}\ \text{C} \] \[ \vec{V}=(2\hat{i}+4\hat{j})\ \text{m s}^{-1} \] \[ \vec{B}=B_0(\hat{i}+4\hat{j})\ \text{T} \]

Step 2: Evaluate the cross product \(\vec{V}\times\vec{B}\).
\[ \vec{V}\times\vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ B_0 & 4B_0 & 0 \end{vmatrix} \] Expanding the determinant, \[ \vec{V}\times\vec{B} = \hat{k} \left(2\cdot4B_0-4\cdot B_0\right) \] \[ = \hat{k}(8B_0-4B_0) \] \[ = 4B_0\hat{k} \]

Step 3: Determine the magnetic force.
Using \[ \vec{F}=q(\vec{V}\times\vec{B}), \] we get \[ \vec{F} = (1.0\times10^{-16})(4B_0)\hat{k} \] \[ = 4B_0\times10^{-16}\hat{k}\ \text{N} \]

Step 4: Compare with the given force.
Given, \[ \vec{F} = 3\times10^{-16}\hat{k}\ \text{N} \] Therefore, \[ 4B_0\times10^{-16} = 3\times10^{-16} \] Cancelling \(10^{-16}\), \[ 4B_0=3 \] \[ B_0=\frac{3}{4} \] \[ B_0=0.75\ \text{T} \]

Step 5: Final conclusion.
Hence, the magnitude of \(B_0\) is \[ \boxed{0.75\ \text{T}} \]