Question

A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field \(\vec{B}=(2\vec{i}+3\vec j-4\vec k)T\). The magnitude of the magnetic force acting on the wire is:

• A. \(\sqrt5 IL\)
• B. 5 IL
• C. \(\sqrt3 IL\)
• D. 3 IL

Answer 1

The Correct Option is B

The magnetic force \( \vec{F} \) on a wire carrying a current \( I \) of length \( L \) in a magnetic field \( \vec{B} \) is given by the formula: \[ \vec{F} = I (\vec{L} \times \vec{B}) \] where \( \vec{L} \) is the vector representation of the wire's length and direction. For a wire along the positive x-axis, we have \( \vec{L} = L\vec{i} \). Given the magnetic field \( \vec{B} = 2\vec{i} + 3\vec{j} - 4\vec{k} \), we begin by computing the cross product \( \vec{L} \times \vec{B} \): \[ \vec{L} \times \vec{B} = (L\vec{i}) \times (2\vec{i} + 3\vec{j} - 4\vec{k}) \] \[ = L (\vec{i} \times 2\vec{i} + \vec{i} \times 3\vec{j} + \vec{i} \times (-4\vec{k})) \] Since \( \vec{i} \times \vec{i} = 0 \), the remaining cross products are: \[ \vec{i} \times \vec{j} = \vec{k}, \quad \vec{i} \times \vec{k} = -\vec{j} \] Thus, \[ \vec{L} \times \vec{B} = L (0 + 3\vec{k} + 4\vec{j}) = L (4\vec{j} + 3\vec{k}) \] Now calculate the magnitude of the force \( \vec{F} \): \[ |\vec{F}| = I | \vec{L} \times \vec{B} | = I \sqrt{(4L)^2 + (3L)^2} \] \[ = I \sqrt{16L^2 + 9L^2} = I \sqrt{25L^2} = 5IL \] Therefore, the magnitude of the magnetic force acting on the wire is \( 5IL \). 

Answer 2

The correct option is (B): 5 IL

\(\vec{F}=I(\vec{l}\times\vec{B})\)

\(I[(L\hat{i})\times(2\hat i+3\hat j-4\hat k)]\)

\(=I(4L\hat j+ 3L\hat k)\)

\(|\vec{F}|=5\,IL\)