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A coil in the shape of an equilateral triangle of side 0.02m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of 5*10-2 T. Calculate the couple acting on the coil when a current of 0.1A is passed through it and the magnetic field is parallel to the plane.

Aditi Chauhan , 12 Years ago
Grade 10
anser 1 Answers
Askiitians Tutor Team

To determine the couple acting on the coil when a current flows through it in a magnetic field, we can use the principles of electromagnetism. The couple, or torque, on a current-carrying coil in a magnetic field can be calculated using the formula:

Torque Calculation

The torque (\( \tau \)) acting on a coil in a magnetic field is given by the equation:

τ = n * B * I * A * sin(θ)

  • n = number of turns in the coil (for a single loop, n = 1)
  • B = magnetic field strength (in teslas)
  • I = current flowing through the coil (in amperes)
  • A = area of the coil (in square meters)
  • θ = angle between the magnetic field and the normal to the plane of the coil

Step 1: Determine the Area of the Coil

The coil is in the shape of an equilateral triangle. The area (\( A \)) of an equilateral triangle can be calculated using the formula:

A = (sqrt(3)/4) * a^2

where \( a \) is the length of a side of the triangle. Given that \( a = 0.02 \, m \):

A = (sqrt(3)/4) * (0.02)^2

A ≈ 0.0001732 \, m^2

Step 2: Identify the Values

Now, we can plug in the values into the torque formula:

  • n = 1 (since it’s a single loop)
  • B = 5 * 10^-2 T
  • I = 0.1 A
  • A ≈ 0.0001732 m²
  • θ = 90° (since the magnetic field is parallel to the plane of the coil, the angle between the magnetic field and the normal is 90°, making sin(θ) = 1)

Step 3: Calculate the Torque

Substituting the values into the torque equation:

τ = 1 * (5 * 10^-2) * (0.1) * (0.0001732) * sin(90°)

τ = 1 * (5 * 10^-2) * (0.1) * (0.0001732) * 1

τ = 8.66 * 10^-6 \, N \cdot m

Final Result

The couple acting on the coil when a current of 0.1 A is passed through it in a magnetic field of 5 * 10^-2 T is approximately 8.66 * 10^-6 N·m. This torque will cause the coil to rotate in the magnetic field, demonstrating the interaction between electricity and magnetism.

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