Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectromagnetic Induction
A closed coil with 100 turns is rotated in a uniform magnetic field B = 4.0 10⁻⁴ T about a diameter that is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm ^2 and its resistance is 4.0 . Determine the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, the average emf in a full turn, and t
Options
- A4.0 10⁻³ V , zero, 1.0 10⁻⁴ C
- B2.0 10⁻³ V , zero, 5.0 10⁻⁵ C
- C1.0 10⁻³ V , zero, 2.5 10⁻⁵ C
- D2.0 10⁻³ V , 2.0 10⁻³ V , 5.0 10⁻⁵ C
Correct answer
B. 2.0 10⁻³ V , zero, 5.0 10⁻⁵ C
Step-by-step solution
Given: Number of turns, N = 100 Magnetic field, B = 4.0 10⁻⁴ T Area of the coil, A = 25 cm ^2 = 25 10⁻⁴ m ^2 Resistance, R = 4.0 Angular velocity, = 300 rpm = 300 2 60 rad/s = 10 rad/s When the coil is perpendicular to the magnetic field, the initial magnetic flux is: _i = NBA (0^ ) = NBA After half a turn (rotation by 180^ ), the final magnetic flux is: _f = NBA (180^ ) = -NBA Change in magnetic flux during half a turn: = | _f - _i| = 2NBA = 2 100 4.0 10⁻⁴ 25 10⁻⁴ = 2.0 10⁻⁴ Wb Time taken for half a turn: t = = 10