Q25. A 300 turns rectangular coil of length 20 cm and breadth 12 cm is carrying a current of 12 A, w

Q: What is the correct answer for CUET UG Physics Q25?

The correct answer is b. The torque (τ) acting on a current-carrying coil in a uniform magnetic field is given by the formula τ = NIABsinθ, where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the magnetic field vector (B) and the normal to the plane of the coil. Given N = 300 turns, I = 12 A, B = 6 T. The dimensions of the coil are lengt

Q: Q25. A 300 turns rectangular coil of length 20 cm and breadth 12 cm is carrying a current of 12 A, when placed in a magnetic field of 6 T. The plane of the coil is making an angle of 60° with the…

The correct answer is b. The torque (τ) acting on a current-carrying coil in a uniform magnetic field is given by the formula τ = NIABsinθ, where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the magnetic field vector (B) and the normal to the plane of the coil. Given N = 300 turns, I = 12 A, B = 6 T. The dimensions of the coil are lengt

Q: What are the options for this CUET UG Physics question?

A. 518.4 N m · B. 259.2 N m · C. 447.8 N m · D. 366.6 N m

Question

Q25. A 300 turns rectangular coil of length 20 cm and breadth 12 cm is carrying a current of 12 A, when placed in a magnetic field of 6 T. The plane of the coil is making an angle of 60° with the magnetic field. What is the torque acting on the coil?

FirstInTest · Accepted Answer

The torque (τ) acting on a current-carrying coil in a uniform magnetic field is given by the formula τ = NIABsinθ, where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the magnetic field vector (B) and the normal to the plane of the coil. Given N = 300 turns, I = 12 A, B = 6 T. The dimensions of the coil are length = 20 cm = 0.20 m and breadth = 12 cm = 0.12 m, so the area A = 0.20 m × 0.12 m = 0.024 m². The plane of the coil makes an angle of 60° with the magnetic field, meaning the angle θ between the normal to the plane and the magnetic field is 90° - 60° = 30°. Now, substitute the values: τ = 300 × 12 A × 0.024 m² × 6 T × sin(30°) = 300 × 12 × 0.024 × 6 × 0.5 = 259.2 N m. Therefore, the torque acting on the coil is 259.2 N m.

Answer

518.4 N m

Answer

259.2 N m

Answer

447.8 N m

Answer

366.6 N m

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