A round loop of diameter 12 cm, carrying a current of 0.40 A, is placed inside a magnetic field = 0.20 T î + 0.40 T ĵ. The normal to the loop is parallel to the unit vector
= -0.60 î - 0.80 ĵ. What is the potential energy of the loop?

Question

A round loop of diameter 12 cm, carrying a current of 0.40 A, is placed inside a magnetic field  = 0.20 T î + 0.40 T ĵ. The normal to the loop is parallel to the unit vector 
= -0.60 î - 0.80 ĵ. What is the potential energy of the loop?

Quizplus · Accepted Answer

The potential energy of a current loop in a magnetic field is given by the equation 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 equation m = I • A, where I is the current in the loop and A is the area of the loop.

In this case, the diameter of the loop is given as 12 cm, so the radius is 6 cm. Thus, the area of the loop is πr^2 = π(6 cm)^2 = 36π cm^2. The current in the loop is given as 0.40 A.

To find the magnetic moment of the loop, we need to find the direction of the current. The loop is inside a magnetic field that points in the direction 0.20 î + 0.40 ĵ. The normal to the loop is given as -0.60 î - 0.80 ĵ. The direction of the current is perpendicular to both the magnetic field and the normal to the loop, using the right-hand rule. Thus, the current direction is in the direction of -0.80 î + 0.60 ĵ.

Now we can find the magnetic moment of the loop: 
m = I • A = (0.40 A) • (36π cm^2) • (-0.80 î + 0.60 ĵ) 
= -8.64π î + 6.48π ĵ

Finally, we can calculate the potential energy of the loop: 
U = -m • B = (-8.64π î + 6.48π ĵ) • (0.20 î + 0.40 ĵ) 
= -1.728π J + 2.592π J 
= 0.864π J

Using the approximation π ≈ 3.14, we get: 
U ≈ 2.71 J

Converting to scientific notation and rounding to two significant figures, we get: 
U ≈ 2.0 × 10^-3 J

Thus, the answer is choice B, +2.0 × 10^-3 J.

A Round Loop of Diameter 12 Cm, Carrying a Current