Question

A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.

a) Calculate the magnitude of its magnetic moment
b) Suppose a uniform magnetic field of magnitude of 0.350 T isapplied parallel to the plane of the loop. What is the magnitude ofthe torque acting on the loop?

Answer 1

a) The magnitude of the magnetic moment is 0.0172 A·m². b) The magnitude of the torque on the loop is 0 N·m.

Step-by-step explanation:

a) To calculate the magnitude of the magnetic moment, we can use the formula μ = nIA, where μ is the magnetic moment, n is the number of turns, I is the current, and A is the area of the coil. In this case, n = 25, I = 15.0 mA = 0.015 A, and A = (5.40 cm)(8.50 cm) = 45.9 cm² = 0.00459 m². Plugging these values into the formula, we get μ = (25)(0.015 A)(0.00459 m²) = 0.0172 A·m².

b) To calculate the magnitude of the torque, we can use the formula τ = μBsinθ, where τ is the torque, μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between μ and B. In this case, μ = 0.0172 A·m² (calculated in part a),

B = 0.350 T, and since the magnetic field is applied parallel to the plane of the loop, θ = 0°.

Plugging these values into the formula, we get τ = (0.0172 A·m²)(0.350 T)(sin 0°)

= 0 N·m.

Answer 2

(a) Magnetic moment will be
`17.212* 10^(-4)A-m^2`

(b) Torque will be
`6.024* 10^(-4)N-m`

Step-by-step explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil
`A=5.4* 8.5=45.9cm^2=45.9* 10^(-4)m^2`

Current is given as
`i=15mA=15* 10^(-3)A`

Number of turns N = 25

(A) We know that magnetic moment is given by
`magnetic\ moment=NiA=25* 45.9* 10^(-4)* 15* 10^(-3)=17.212* 10^(-4)A-m^2`

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by
`\tau =BINA=0.350* 15* 10^(-3)* 25* 45.9* 10^(-4)=6.024* 10^(-4)N-m`