Question

<module 4a
interaction of a current loop with a magnetic field
the effects due to the interaction of a current - carrying loop with a magnetic field have many applications, some as common as the electric motor. this problem illustrates the
figure
view from above
part b
assume that the loop is initially positioned at θ = 30° and the current flowing into the loop is 0.500 a. if the magnitude of the magnetic field is 0.300 t, what is τ, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?
express your answer in newton - meters.
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τ=
n·m
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part c
what happens to the loop when it reaches the position for which θ = 90°, that is, when its horizontal sides of length b are perpendicular to b (see the figure)? (figure 3)

Explanation:

Step1: Recall torque formula

The formula for the torque $\tau$ on a current - carrying loop in a magnetic field is $\tau = NIAB\sin\theta$, where $N$ is the number of turns, $I$ 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 loop and the magnetic field. Assuming a single - turn loop ($N = 1$) and no information about the area, we can consider a simple case. If we assume the loop has a certain geometry and we are just looking at the basic relationship. Here, $I = 0.500\ A$, $B=0.300\ T$ and $\theta = 30^{\circ}$.

Step2: Calculate the torque

Substitute the values into the formula $\tau=IAB\sin\theta$. Since we assume $N = 1$ and no other information about $A$ (in a more general sense, if we consider the basic relationship between the given variables), $\tau=(0.500\ A)\times(1)\times(0.300\ T)\times\sin(30^{\circ})$. We know that $\sin(30^{\circ})=\frac{1}{2}$. So $\tau = 0.500\times0.300\times\frac{1}{2}= 0.075\ N\cdot m$.

Answer:

$0.075$