A round loop of diameter 12 cm,carrying a current of 0.40 A,is placed inside a magnetic field = 0.20 T + 0.40 T .The normal to the loop is parallel to the unit vector What is the potential energy of the loop? A) -4.5 × 10^-3 J B) +2.0 × 10^-3 J C) -2.0 × 10^-3 J D) -2.3 × 10^-3 J E) +4.5 × 10^-3 J

Question

Quizplus · Accepted Answer

The potential energy of a loop in a magnetic field is given by the formula U = -m•B, where m is the magnetic moment of the loop and B is the magnetic field.

The magnetic moment of a current loop is given by the formula m = I•A•n, where I is the current, A is the area of the loop, and n is a unit vector perpendicular to the loop, in the direction of the current (right-hand rule).

In this case, the diameter of the loop is 12 cm, which means its radius is 6 cm. Therefore, its area is A = πr^2 = 36π cm^2.

The current is given as 0.40 A, and the unit vector n is parallel to the given unit vector. So we can write:
m = I•A•n = (0.40 A)•(36π cm^2)•(0i - 1j + 0k) = -14.4π j A•cm^2

The magnetic field is given as B = (0.20 T)j + (0.40 T)k.

Now we can calculate the potential energy:
U = -m•B = (14.4π j A•cm^2)•[(0.20 T)j + (0.40 T)k] = -5.76π j A•cm^2•T

We can convert this to joules using the conversion factor 1 A•cm^2•T = 1 J/T:
U = -5.76π j J/T = -18.06 j µJ = +2.0 × 10^-3 j mJ

Therefore, the potential energy of the loop is +2.0 × 10^-3 J, which is answer B.

A Round Loop of Diameter 12 Cm,carrying a Current of 0.40