Question

A current of 17.0 mA is maintained in a single circular loop with a circumference of 2.00 m. A magnetic field of 0.800 T is directed parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop. (b) What is the magnitude of the torque exerted on the loop by the magnetic field?

Answer 1

Answer:

(A) $magnetic\ moment=5.41\times 10^{-3}A-m^2$

(B) Torque is given as $T=MB=0.8\times 5.41\times 10^{-3}=4.32\times 10^{-3}N-m$

Explanation:

We have given current in the circular loop $i=17mA=17\times 10^{-3}A$

Circumference = 2 m

So $2\pi r=2$

$r=\frac{2}{2\times 3.14}=0.3184m$

Area $A=\pi r^2=3.14\times 0.3184^2=0.3183m^2$

(A) Magnetic moment is given by

$M=current\times area=17\times 10^{-3}\times 0.3183=5.411\times 10^{-3}A-m^2$

(b) Magnetic field is given B = 0.8 T

Torque is given by  $T=MB=0.8\times 5.41\times 10^{-3}=4.32\times 10^{-3}N-m$

Answer 2

Part A: The moment of the loop is $5.4 \times 10^{-3} Am^2$.

Part B: The torque exerted on the loop by the magnetic field is $4.32 \times 10^{-3} \;\rm Nm$.

What is torque?

Torque is defined as the force acting on an object causes that object to rotate.

Given that the current I in the circular loop is 17.0 mA and the circumference of the loop is 2.00 m.

The radius of the circular loop is calculated as given below.

$2\pi r= 2$

$r = \dfrac {2}{2\times 3.14}$

$r = 0.318\;\rm m$

Part A:

The moment in the circular loop is calculated as given below.

$M = i A$

$M = 17 \times 10^{-3}\times 3.14 \times0.318^2$

$M = 5.4 \times 10^{-3} Am^2$

Hence the moment of the loop is $5.4 \times 10^{-3} Am^2$.

Part B:

The torque exerted on the loop by the magnetic field is given below.

$T = MB$

Where T is the torque and B is the magnetic field.

$T = 5.4 \times 10^{-3} \times 0.800$

$T = 4.32 \times 10^{-3} \;\rm Nm$

Hence the torque exerted on the loop by the magnetic field is $4.32 \times 10^{-3} \;\rm Nm$.

To know more about the torque, follow the link given below.

https://brainly.com/question/6855614.