A coil of 50 turns is normally placed in a magnetic field of 0.6T.the area of this coil is 0.2 m² and the resistance in the circuit is 10 ohm.calculate induced charge when: (a) coil is rotated by 180°(b) coil is removed from the magnetic field.
Answers
Answer:
GIVEN:-
=>Number of turns =50 magnetic field.
=>(E) = 0.6T area (A)=0.2m^2.
=>Resistance (R)=10Ω.
TO FIND:-
(a) Coil is rotated by 180°.
(b) Coil is removed from the magnetic field.
UNDERSTANDING THE CONCEPT:-
According to the question,
=>The magnetic flux is passing through the coil in initial state,
(ϕ1) = BAcosθ
=BAcos0=BA
=0.6 × 0.2 × 1
=0.12Wb
a) The magnetic flux, after rotating the angle of 180°.
=> ϕ2=BAcos(1800)
=> −0.6×0.2
=> 0.12Wb
∴ Induced Charge
q = N/ R(ϕ1 − ϕ2 )
=> 50/10 (0.12−(−0.12)
= 50/10 ×0.24
= 1.20C
Now we have arrived with the answer for the coil rotated by 180°.
b) The induced charge,
when the coil comes out from the magnetic filed (ϕ^2 )=0.
=>50/10 (0.12)
=>0.60C
Therefore, now we have arrived with the answer for the coil is removed from the magnetic field.
(a) Coil is rotated by 180°is 1.20C
(b) Coil is removed from the magnetic ffield is 0.60C