Question

You have a wire of length L = 1.9 m for making a square coil of a dc motor. The current in the coil is I = 2.1 A, and the magnetic field of the motor has a magnitude of B = 0.30 T. Find the maximum torque exerted on the coil when the wire is used to make

(a) a single-turn square coil and
(b) a two-turn square coil.

Answer 1

Answer: a) 0.14213 Nm b) 0.0711 Nm

Step-by-step explanation:

Assume that magnetic field is perpendicular to the area

a) a single-turn square coil

L = 1.9 m

thus L/4 = 1.9 m/ 4 = 0.475 m ; Area = (0.475 m)² = 0.2256 m²

I = 2.1 A

B = 0.30 T

N = 1

θ = 90⁰

maximum torque τ = N IA B Sinθ = 1 * 2.1 * 0.3*0.2256* Sin90⁰ = 0.14213 Nm

b) a two-turn square coil

L = 1.9 m

thus L/4 = 1.9 m/ 8 = 0.2375 m ; Area = (0.2375 m)² = 0.0564 m²

I = 2.1 A

B = 0.30 T

N = 2

θ = 90⁰

maximum torque τ = N IA B Sinθ = 2 * 2.1 * 0.3*0.0564* Sin90⁰ = 0.0711 Nm

Answer 2

(A) 0.1421 N-m

(b) 0.2842 N-m

Step-by-step explanation:

We have given length of the wire L =1.9 m

So side of the square
`=(1.9)/(4)=0.475m`

We know that area of the square
`=side^2=0.475^2=0.2256m^2`

Current in the loop i = 2.1 A

Magnetic field B = 0.3 T

(A) Number of turns N = 1

We know that torque is given by

`\tau =BINAsin\Theta`

For maximum torque
`sin\Theta =1`

So torque
`\tau =1* 2.1* 0.3* 0.2256=0.1421N-m`

(b) Number of turns N = 2

We know that torque is given by

`\tau =BINAsin\Theta`

For maximum torque
`sin\Theta =1`

So torque
`\tau =2* 2.1* 0.3* 0.2256=0.2842N-m`