Answer:
Explanation:
Given that:
The radius = 10 cm
No of turns = 125
Current in the coil (I) = 0.250 A
Uniform magnetic field B = 2.50 mT = 2.5 × 10⁻³ T
a) The sketch of the situation can be seen below
b)The magnetic moment of the coil is [tex]\mu^{\to} = ni A^{\to}[/tex]
where the direction of the magnetic moment is the same and equivalent to the direction of the area vector in the diagram. The region vector of the coil is parallel to the magnetic field.
In this manner, the direction of magnetic moment is perpendicular to the current loop by right-hand rule direction; which implies out of the page.
(c)
The torque [tex]\tau = \mu^{\to} \times B^{\to}[/tex]
[tex]= \mu B sin \theta[/tex]
where;
[tex]\theta[/tex] between [tex]\mu[/tex] and B = 0
Thus;
[tex]\tau = \mu B sin (0)[/tex]
[tex]\tau = 0[/tex]
