Question

A wire carrying current 'I' along x axis has length 'L' and it is kept in a magnetic field \(B(\hat{i}+2\hat{j}-2\hat{k})\) T. The magnitude of magnetic force acting on the wire is

• A. \(\sqrt{8}\,\text{ILB}\)
• B. \(2\,\text{ILB}\)
• C. \(4\,\text{ILB}\)
• D. \(\sqrt{2}\,\text{ILB}\)

Hint:

Force is I L cross B; only the components of B perpendicular to the wire matter.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The force on a straight wire is \(\vec F = I\,\vec L\times\vec B\). The wire is along the \(x\)-axis, so \(\vec L = L\hat i\).

Step 2: Cross product:
\[ \hat i\times(\hat i + 2\hat j - 2\hat k) = 0 + 2\hat k + 2\hat j = 2\hat j + 2\hat k \]
(using \(\hat i\times\hat j = \hat k\) and \(\hat i\times\hat k = -\hat j\)).

Step 3: Magnitude:
\[ |\vec F| = ILB\sqrt{2^2 + 2^2} = \sqrt8\,ILB \]

Step 4: Check:
Option (A). The component of \(\vec B\) along the wire (the \(\hat i\) part) gives no force. Option (C) 4ILB would be the sum of the two perpendicular components added directly, but they are in different directions.

Final Answer:
Only the 2j - 2k part of B matters, giving root 8 ILB. \[ \boxed{\text{(A) }\sqrt8\,ILB} \]