A square loop with a side length of $1 ext{ m}$ and resistance of $1 \Omega$ is placed in a uniform magnetic

Question

A square loop with a side length of $1 	ext{ m}$ and resistance of $1 \Omega$ is placed in a uniform magnetic field of $0.5 	ext{ T}$. The plane of the loop is perpendicular to the direction of the magnetic field. The magnetic flux through the loop is:

Accepted Answer

The magnetic flux $\Phi_B$ through a surface is defined as the scalar product of the magnetic field vector $\mathbf{B}$ and the area vector $\mathbf{A}$:
$$\Phi_B = \mathbf{B} \cdot \mathbf{A} = BA \cos 	heta$$

1.  **Identify Given Values:**
    *   Magnetic Field ($B$) = $0.5 	ext{ T}$.
    *   Side length of square ($l$) = $1 	ext{ m}$.
    *   Area of loop ($A$) = $l^2 = (1 	ext{ m})^2 = 1 	ext{ m}^2$.
    *   Resistance ($R$) = $1 \Omega$ (This is distractor information not needed for flux calculation).

2.  **Determine Angle ($	heta$):**
    *   The problem states the *plane* of the loop is perpendicular to the magnetic field.
    *   The area vector ($\mathbf{A}$) is always normal (perpendicular) to the plane of the loop.
    *   Therefore, the angle $	heta$ between the magnetic field $\mathbf{B}$ and the area vector $\mathbf{A}$ is $0^\circ$.

3.  **Calculate Flux:**
    $$\Phi_B = (0.5 	ext{ T})(1 	ext{ m}^2) \cos(0^\circ)$$
    $$\Phi_B = 0.5 	imes 1 	imes 1 = 0.5 	ext{ Wb}$$

Thus, the magnetic flux through the loop is $0.5 	ext{ Wb}$.

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