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A flat 145-turn current-carrying loop is immersed in a uniform magnetic field. The area of the loop is 6.75 cm2 and the angle between its magnetic dipole moment and the field is 31.5°. Find the strength of the magnetic field that causes a torque of 1.33 × 10-5 N·m to act on the loop when a current of 2.11 mA flows in it.

Respuesta :

Answer:

Magnetic field, B = 0.123 T

Explanation:

It is given that,

Number of turns in current carrying loop, N = 145

Area of the loop, [tex]A=6.75\ cm^2=0.000675\ m^2[/tex]

The angle between its magnetic dipole moment and the field is 31.5°

Torque, [tex]\tau=1.33\times 10^{-5}\ N.m[/tex]

Current, [tex]I=2.11\ mA=2.11\times 10^{-3}\ A[/tex]

Toque on the current carrying loop is given by :

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

[tex]B=\dfrac{\tau}{INA\ sin\theta}[/tex]

[tex]B=\dfrac{1.33\times 10^{-5}}{2.11\times 10^{-3}\times 145\times 0.000675\ sin(31.5)}[/tex]

B = 0.123 T

So, the strength of the magnetic field is 0.123 T. Hence, this is the required solution.

onyebuchinnaji

The magnetic field strength created around the loop is 0.123 T.

The given parameters;

  • number of turns of the wire, N = 145 turns
  • area of the loop = 6.75 cm²
  • angle between its magnetic dipole moment and the field, θ = 31.5
  • torque, τ = 1.33 x 10⁻⁵ N.m
  • current on the loop, I = 2.11 mA.

The torque on the loop is calculated as follows;

[tex]\tau = NBIA sin(\theta)\\\\B= \frac{\tau}{NIA sin(\theta)}[/tex]

Where;

B is the magnetic field strength

[tex]B= \frac{\tau}{NIA sin(\theta)} \\\\B = \frac{(1.33 \times 10^{-5} )}{145 \times (2.11\times 10^{-3}) \times (6.75 \times 10^{-4})\times sin(31.5)} \\\\B = 0.123 \ T[/tex]

Thus, the magnetic field strength created around the loop is 0.123 T.

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