Question

A wire has a length of 7.00 \times 10^{-2} m and is used to make a circular coil of one turn. There is a current of 4.30 A in the wire. In the presence of a 2.50-T magnetic field, what is the maximum torque that this coil can experience

Answer 1

Answer:

Maximum torque that coil experience is 4.19 x 10⁻³ N-m

Explanation:

Given :

Length of wire, L = 7 x 10⁻² m

Current flowing through the circular coil, I = 4.30 A

Magnetic field applied, B = 2.50 T

Number of turns, N = 1

The relation to determine torque experienced by the circular coil due to magnetic field:

τ = NIAB sinθ   ....(1)

Here A is the area of the circular coil and θ is the angle between circular coil and magnetic field.

Consider the radius of the circular coil be r.

But, length of wire = Circumference of the circular coil

So, L = 2πr

r = L/2π    ....(2)

Area of circular coil, A = πr²

Substitute equation (2) in the above equation.

$A=\frac{L^{2} }{4\pi }$

Substitute the above equation in equation (1)

$\tau=\frac{NIBL^{2} \sin \theta}{4\pi }$

The torque will be maximum when sin θ = 1, that is, θ = 90⁰. Thus, the above equation becomes:

$\tau=\frac{NIBL^{2}}{4\pi }$

Substitute the suitable values in the above equation.

$\tau=\frac{1\times4.30\times2.50\times(7\times10^{-2}) ^{2}}{4\pi }$

τ = 4.19 x 10⁻³ N-m

Answer 2

The torque on the coil is directly proportional to the current of the wire. The maximum torque that a given coil is experiencing is $4.19 \times 10^{-3} \rm \ Nm$.

The maximum torque:

$\tau = \dfrac {NIBL^2}{4\pi}$

Where,

$N$ - numbers of turn  = 1

$I$ - current  4.3 A

$B$ - magnetic field = 2.5 T

$L$ - length of wire = $7\times 10^{-2}$ m

Put the values in the equation,

$\tau = \dfrac {1 \times 2.5 \times 4.3 \times( 7\times 10^{-2})^2}{2\pi}\\\\\tau = 4.19 \times 10^{-3} \rm \ Nm$

Therefore, the maximum torque that a given coil is experiencing is $4.19 \times 10^{-3} \rm \ Nm$.

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