AP EAPCET 07-Jul-2022 Shift 1 Paper

Due to solenoid a magnetic filled exists around coil placed at Centre

Magnetic field intensity produced by solenoid,
B=µ0nI
Flux linked with coil is
φB=N⋅B⋅A,
Where, N= Number of turns in coil
A=Area‌of‌coil
As current in solenoid is reversed, change in flux is
∆φB=2N.B.A
This change in flux produces an emf in coil given by
Induced emf e= Rate of change of flux
⇒e=

2NBA

∆t

=

2N(µ0nI)A

∆t

Here N=200,A=25cm2=25×10−4m2
I=

4

π

A,n=100‌tuns∕cm=100×100‌turns∕m
µ0=4π×10−7T⋅mA−1 and ∆t=0.04 s
So induced emf :
e=

2×200×4π×10−7×104× 4 π ×25×10−4

0.04

=0.4V
Induced emf in coil is 0.4V