Question

A square loop and a rectangular loop are each made from the same length of wire. Each loop contains a single turn, and carries the same amount of current. The loops are placed in the same uniform magnetic field. The long sides of the rectangle are five times as long as its short sides. Determine the ratio of the maximum torque on the square loop to the maximum torque on the rectangular loop. (Enter your answer as a fraction, e.g., "5/6".)

Answer 1

Answer:

9/5

Explanation:

The torque due to magnetic field is given by $\vec{\tau} = I\vec{A} \times \vec{B}=IABsin(\theta)$, where $\theta$ is the angle between the magnetic moment vector ($I\vec{A}$) and the magnetic field ($\vec{B}$). Note that B is constant in magnitude and direction, therefore, we can obtain the ratio of the torques by finding the ratio of the areas for both loops:

We start by naming the total length of the wires $L$

1) Square loop:

the square sides are $\frac{L}{4}$ each. Therefore its are will be $A_{square} =(\frac{L}{4})^{2}=\frac{L^2}{16}$

1) Rectangle loop:

now the rectangle has also a perimeter of $L$. Let $x$ be the length of the short sides of it, then we have:

$L = 5x+5x+x+x=12x=>x=\frac{L}{12}$

now the area of the rectangle is $A_{rectangle} =\frac{L}{12}*\frac{5L}{12}=\frac{5L^2}{144}$

then the ratio $\frac{A_{square}}{A_{rectangle}}=\frac{\frac{L^2}{16}}{\frac{5L^2}{144}}=\frac{9}{5} = \tau_{ratio}$