Magnetism And Matter

The horizontal component of the Earth's magnetic field is determined by resolving the total magnetic field vector B into its horizontal and vertical components. This is done by considering the angle that the total magnetic field B makes with the horizontal.

To find the horizontal component B_H, we use the cosine of the angle since the horizontal component is adjacent to the angle in the right triangle formed by the magnetic field vector. This gives us the equation: B_H = Bcos( )

This equation shows that the horizontal component is the product of the total magnetic field B and the cosine of the angle . Therefore, the correct representation of the horizontal component of the Earth's magnetic field is: B_H = Bcos( )

To determine which wire loop configuration experiences greater torque when placed in a uniform magnetic field with an electric current flowing through them, we need to consider the formula for torque (τ) on a current-carrying loop in a magnetic field:

τ = nIAB sin(θ)

Where:

• n = number of turns of the loop (which is 1 in this case since it's a single loop)
• I = current passing through the loop
• A = area of the loop
• B = magnetic field strength
• θ = angle between the normal to the loop and the magnetic field (assuming θ=90° for maximum torque, so sin(θ)=1)

Given that the lengths of the wires are equal, we need to find the configuration that maximizes the area (A) since all other factors are constant.

Calculating Areas:

• Circle: Given circumference = length = L, the radius r of the circle is L/2π. The area of the circle is:
  A_circle = πr² = π(L/2π)² = L²/4π
• Semi-circle: When the wire is a semi-circle, half the perimeter is used to form a circle with radius r = L/π. The area is half that of a circle with this radius.
  A_semicircle = (1/2)×π(r²) = π(L/π)²/2 = L²/2π
• Square: Each side of the square is L/4. Hence, the area is:
  A_square = (L/4)² = L²/16

Comparing the areas:

• A_circle = L²/4π
• A_semicircle = L²/2π
• A_square = L²/16

Since 4π ≈ 12.57, A_circle > A_semicircle > A_square. Therefore, the circular loop experiences the greatest torque.

The correct answer is: Circle