Question

A coil of 40amp resistance has 100 turns and radius is 6mm is connected to ammeter of resistance of 160 ohms and the coil is placed in perpendicular to the magnetic field. When the coil is taken out of the field ,32micro columbs charge flows through it. The magnetic induction field strength is​

Answer 1

Explanation:

d Φ/dt=e

rate of change of fluex =potential diffetence

e=IR

Φ=BAN

BAN=IR

B=IR/AN

GIVEN

I=32×10^-6

R=160 ohm

A=PIE×r^2

A×N=

0.011304

B=32×10^-6×160/0.011304

B=453097.34×10^-6

B=0.4530 T