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.
Correct Answer: 160
Approach Solution - 1
Given: \( \ell = 2 \, \text{m} \) and \( i = 2 \, \text{A} \)
Magnetic field values are:
\( B(x = 0) = B_0 \) and \( B(x = 2) = 9B_0 \)
We know that the magnetic force on a current-carrying conductor is given by:
\( F = i \ell B \)
Therefore, for the two sides of the loop:
\( F_1 = i \ell B_0 \) and \( F_2 = 9i \ell B_0 \)
The net force acting on the loop is:
\( F = F_2 - F_1 = 8i \ell B_0 \)
Substituting the given values:
\( F = 8 \times 2 \times 2 \times 5 = 160 \, \text{N} \)
Hence, the net force on the loop is:
\( F = 160 \, \text{N} \)
Approach Solution -2
Step 1: Magnetic force on a current-carrying conductor The magnetic force on a straight conductor in a magnetic field is given by:
\[ \vec{F} = I \int (\vec{dl} \times \vec{B}), \]
where \( I \) is the current, \( \vec{dl} \) is the element of the wire, and \( \vec{B} \) is the magnetic field.
Step 2: Magnetic field variation The magnetic field is given as:
\[ \vec{B} = B_0 (1 + 4x) \hat{k}. \]
Since the loop is aligned parallel to the \( x\text{-}y \)-plane, \( \vec{B} \) varies with \( x \), causing forces on the opposite sides of the loop to not cancel.
Step 3: Net force on the loop For the two vertical sides of the loop (along the \( y \)-direction):
\[ F_\text{vertical} = I L \Delta B, \]
where \( \Delta B \) is the difference in the magnetic field at \( x = 2 \, \text{m} \) and \( x = 0 \, \text{m} \).
\[ \Delta B = B_0 (1 + 4 \cdot 2) - B_0 (1 + 4 \cdot 0), \]
\[ \Delta B = 5 \cdot (1 + 8) - 5 \cdot (1 + 0), \]
\[ \Delta B = 45 - 5 = 40 \, \text{T}. \]
Substitute \( I = 2 \, \text{A} \), \( L = 2 \, \text{m} \), and \( \Delta B = 40 \, \text{T} \):
\[ F_\text{vertical} = 2 \cdot 2 \cdot 40 = 160 \, \text{N}. \]
For the horizontal sides of the loop (along the \( x \)-direction), the magnetic forces cancel out because the magnetic field is uniform along the \( y \)-axis.
Step 4: Final result The net magnetic force on the loop is: \[ 160 \, \text{N}. \]
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