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:

Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength \( B = 0.5 \, \text{T} \). If the current in the wire is \( I = 2 \, \text{A} \) and the wire makes an angle of \( 30^\circ \) with the magnetic field, find the force per unit length on the wire.

Answer 1

To solve for the magnitude of the magnetic force acting on the wire, we use the formula for the magnetic force on a current-carrying wire, which is:

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

where \(I\) is the current, \(\vec{L}\) is the length vector of the wire, and \(\vec{B}\) is the magnetic field.

Given that the wire is along the positive x-axis, the length vector can be written as:

\(\vec{L} = L\vec{i}\)

The magnetic field is given as:

\(\vec{B} = 2\vec{i} + 3\vec{j} - 4\vec{k}\)

Now, compute the cross product \(\vec{L} \times \vec{B}\):

\(\vec{L} \times \vec{B} = (L\vec{i}) \times (2\vec{i} + 3\vec{j} - 4\vec{k})\)

The cross product of a vector with itself is zero, so:

\((L\vec{i} \times 2\vec{i}) = 0\)

Compute the remaining terms:

\((L\vec{i} \times 3\vec{j}) = 3L(\vec{i} \times \vec{j}) = 3L\vec{k}\)

\((L\vec{i} \times (-4\vec{k})) = -4L(\vec{i} \times \vec{k}) = 4L\vec{j}\)

Combine these results to get:

\(\vec{L} \times \vec{B} = 3L\vec{k} + 4L\vec{j}\)

Thus, the cross product is:

\(\vec{L} \times \vec{B} = (4L\vec{j} + 3L\vec{k})\)

The magnitude of this cross product is given by:

\(|\vec{L} \times \vec{B}| = \sqrt{(4L)^2 + (3L)^2} = \sqrt{16L^2 + 9L^2} = \sqrt{25L^2} = 5L\)

Therefore, the magnitude of the force is:

\(F = I \times |\vec{L} \times \vec{B}| = 5IL\)

Thus, the correct answer is 5 IL.

Answer 2