Question

A square loop of side 30 cm and wire cross section having diameter 4mm is placed perpendicular to a magnetic field changing at the ratio 0.2 T/s. find induced current in the wire loop ​

Answer 1

Answer:

Explanation:

indecd voltage = 0.2 x 0.09

                          =0.018

To find currect we need resistance which is not given in question

Answer 2

A square loop of side 30 cm and wire cross section having diameter 4mm is placed perpendicular to a magnetic field changing at the ratio 0.2 T/s

To find : induced current in the wire loop.

solution : according to Faraday's law,

induced emf in the wire, |ε| = dΦ/dt

= d(BA)/dt

= A dB/dt

= (πr²) dB/dt

resistance of wire, R = ρl/(πr²)

so current in the wire, i = ε/R

= {(πr²)² dB/dt}/ρl

here rate of change of magnetic field = dB/dt = 0.2T/s

radius, r = 2 × 10¯³ m

resistivity of wire material, ρ = 1.23 × 10^-8 Ωm

and length of wire = 4 × 0.3 m

= π² × (2 × 10¯³)² × 0.2/(1.23 × 10^-8 × 4 × 0.3)

= 5.34 × 10² A

Therefore the induced current in the wire loop is 5.34 × 10² A