A flat rectangular loop of wire carrying a $4.0$-A current is placed in a uniform $0.60 - T$ magnetic field. The magnitude of the torque acting on this loop when the plane of the loop makes a $30 ^ { \circ }$ angle with the field is measured to be $1.1 \mathrm {~N} \cdot \mathrm { m }$. What is the area of this loop?
A) $0.40 \mathrm {~m} ^ { 2 }$
B) $0.80 \mathrm {~m} ^ { 2 }$
C) $0.53 \mathrm {~m} ^ { 2 }$
D) $0.26 \mathrm {~m} ^ { 2 }$
E) $0.20 \mathrm {~m} ^ { 2 }$

Question

Quizplus · Accepted Answer

The torque ($\tau$) on a current-carrying loop in a magnetic field is given by $\tau = nIAB\sin(\theta)$, where $n$ is the number of turns (which is 1 in this case), $I$ is the current, $A$ is the area of the loop, $B$ is the magnetic field strength, and $\theta$ is the angle between the normal to the plane of the loop and the magnetic field. Rearranging for $A$, we get $A = \frac{\tau}{IB\sin(\theta)}$. Substituting the given values: $A = \frac{1.1}{4.0 \times 0.60 \times \sin(30^\circ)} = \frac{1.1}{4.0 \times 0.60 \times 0.5} = 0.53 \, \text{m}^2$.

A Flat Rectangular Loop of Wire Carrying A 4.04.04.0 -A Current Is Placed in a Uniform

A Flat Rectangular Loop of Wire Carrying A $4.0$ -A Current Is Placed in a Uniform