The maximum torque on a 135-turn square loop of wire, 18.0 cm on a side, carrying a 47.0 A current in a 1.60 T magnetic field is calculated using the formula for maximum torque on a current-carrying loop. The calculations yield a maximum torque of 4.678 N⋅m.
The question asks to calculate the maximum torque on a square loop of wire carrying current in a magnetic field. The formula to find maximum torque (τ) on a current-carrying loop in a magnetic field is given by τ = nIABsin(θ), where n is the number of turns, I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the normal to the loop and the magnetic field. The maximum torque occurs when sin(θ) = 1, i.e., θ = 90°.
Given that the number of turns (n) is 135, the side of the square loop (l) is 18.0 cm (which needs to be converted to meters to get the area in square meters), the current (I) is 47.0 A, and the magnetic field (B) is 1.60 T, the area (A) of the square loop is l2 = (0.18 m)2. The formula becomes τ = 135 * 47.0 A * (0.18 m)2 * 1.60 T.
To find the maximum torque, calculate:
- Convert the side length from cm to m: 18.0 cm = 0.18 m
- Calculate the area of the square loop: A = l2 = (0.18 m)2 = 0.0324 m2
- Substitute the values into the torque formula: τ = nIAB = 135 * 47.0 A * 0.0324 m2 * 1.60 T = 4.678 N⋅m
Hence, the maximum torque will be 4.678 N⋅m.