Question

A square loop of edge length $2 \, \text{m}$ carrying current of $2 \, \text{A}$ is placed with its edges parallel to the x-y axis. A magnetic field is passing through the x-y plane and expressed as \[ \vec{B} = B_0 (1 + 4x) \hat{k}, \] where $B_0 = 5 \, \text{T}$. The net magnetic force experienced by the loop is ______ N.

Answer 1

Correct Answer: 160

Provided: \( \ell = 2 \, \text{m} \) and \( i = 2 \, \text{A} \)

Magnetic field values are:

\( B(x = 0) = B_0 \) and \( B(x = 2) = 9B_0 \)

The magnetic force on a current-carrying conductor is expressed as:

\( F = i \ell B \)

For the two sides of the loop:

\( F_1 = i \ell B_0 \) and \( F_2 = 9i \ell B_0 \)

The resultant force on the loop is:

\( F = F_2 - F_1 = 8i \ell B_0 \)

Substituting the given values:

\( F = 8 \times 2 \times 2 \times 5 = 160 \, \text{N} \)

Consequently, the net force on the loop is:

\( F = 160 \, \text{N} \)