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 magnetic field is given by τ = nIABsinθ, where:- n = number of turns in the coil- I = current flowing through the coil- A = area of the coil- B = magnetic field strength- θ = angle between the magnetic field vector (B) and the normal to the plane of the coil (area vector A).Given:Number of turns (n) = 300Length of coil (l) = 20 cm = 0

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 magnetic field is given by τ = nIABsinθ, where:- n = number of turns in the coil- I = current flowing through the coil- A = area of the coil- B = magnetic field strength- θ = angle between the magnetic field vector (B) and the normal to the plane of the coil (area vector A).Given:Number of turns (n) = 300Length of coil (l) = 20 cm = 0

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?
(a) 518.4 N m
(b) 259.2 N m
(c) 447.8 N m
(d) 366.6 N m

FirstInTest · Accepted Answer

The torque (τ) acting on a current-carrying coil in a magnetic field is given by τ = nIABsinθ, where:-   n = number of turns in the coil-   I = current flowing through the coil-   A = area of the coil-   B = magnetic field strength-   θ = angle between the magnetic field vector (B) and the normal to the plane of the coil (area vector A).Given:Number of turns (n) = 300Length of coil (l) = 20 cm = 0.20 mBreadth of coil (b) = 12 cm = 0.12 mArea of the coil (A) = l × b = 0.20 m × 0.12 m = 0.024 m²Current (I) = 12 AMagnetic field (B) = 6 TAngle between the plane of the coil and the magnetic field = 60°.Therefore, the angle θ between the normal to the plane of the coil and the magnetic field is θ = 90° - 60° = 30°.Now, calculate the torque:τ = 300 × 12 A × 0.024 m² × 6 T × sin(30°)τ = 300 × 12 × 0.024 × 6 × 0.5τ = 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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