Question

A coil of 50 turns is normally placed in a magnetic field of 0.6 T .the area of this coil is 0.2 m² and the resistance in the circuit is 10ohm.calculate induced charge when: (a) coil is rotated by 180° (b) coil is removed from the magnetic field.​

Answer 1

Explanation:

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

$\huge\bf{Answer:}$

a) The magnetic flux, after rotating the angle of 180°.

=> ϕ2=BAcos(1800)

=> −0.6×0.2

=> 0.12Wb

∴ Induced Charge 

$\bf{q = \dfrac{n}{r} (ϕ1−ϕ2)</p><p></p><p>}$

=> 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.

$\sf{q = \dfrac{n}{r} (ϕ1−ϕ2)</p><p></p><p>}$

=>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