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A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.500 T. If the coil carries a current of 25.0 mA, find the magnitude of the maximum possible torque exerted on the coil.

Respuesta :

Answer:

The maximum torque in the coil is [tex]4.9\times 10^{-3}\ N-m[/tex].

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

[tex]\tau=NIAB\ \sin\theta[/tex]

For maximum torque, [tex]\theta=90^{\circ}[/tex]

[tex]\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m[/tex]

So, the maximum torque in the coil is [tex]4.9\times 10^{-3}\ N-m[/tex].

onyebuchinnaji

The maximum torque exerted on the coil is 0.0049 Nm.

The given parameters;

  • number of turns, N = 50 turn
  • radius of the circular coil, r = 5 cm = 0.05 m
  • magnitude of the magnetic field, B = 0.5 T
  • current in the coil, I = 25 mA

The maximum torque exerted on the coil is calculated as follows;

[tex]\tau = NIAB[/tex]

where;

  • A is the area of the coil

[tex]A = \pi r^2\\\\A = \pi \times (0.05)^2\\\\A = 7.855 \times 10^{-3} \ m^2[/tex]

[tex]\tau = (50) \times (25\times 10^{-3}) \times (7.855\times 10^{-3}) \times (0.5)\\\\\tau = 0.0049 \ N.m[/tex]

Thus, the maximum torque exerted on the coil is 0.0049 Nm.

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