Question

A rigid rectangular loop, which measures 0.30 m by 0.40 m, carries a current of 5.5 A, as shown in the figure. A uniform external magnetic field of magnitude 2.9 T in the negative x direction is present. Segment CD is in the xz-plane and forms a 35° angle with the z-axis, as shown. Find the magnitude of the external torque needed to keep the loop in static equilibrium.

Answer 1

Answer:

The magnitude will be "$(1.097 \ N/m)\hat{j}$". The further explanation is given below.

Explanation:

Trying to determine what is usual for a rectangular loop plane,

⇒  $\hat{n}=(Cos35^{\circ})(-\hat{i})+(Sin35^{\circ})(\hat{k})$

Magnetic moment is given as:

μ = $IA\hat{n}$

On putting the values in the above formula, we get

μ = $(5.5A)[(0.3m)(0.4m)][(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(-\hat{k})]$

  = $0.66 Am^2[(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(\hat{k})]$

Now the external vector-shaped torque seems to be:

Magnitude,

$\sigma=\hat{\mu}\times\hat{\beta}$

On putting the values in the above formula, we get

⇒ $\sigma=[(0.06Am^2)(Cos35^{\circ}(-\hat{i})+Sin35^{\circ}(-\hat{k})]\times (2.9T)(-\hat{i})$

⇒    $= (0.66)(2.9)Cos35^{\circ}(\hat{i}\times \hat{i})+(0.66)(2.9)Sin35^{\circ}(\hat{k}\times \hat{k})$

⇒    $= 0+(1.97 \ N/m)\hat{j}$

⇒    $=(1.097 \ N/m)\hat{j}$