emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
Answer:
import java.util.Scanner;
public class InputExample {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int birthMonth;
int birthYear;
birthMonth = scnr.nextInt();
birthYear = scnr.nextInt();
System.out.println(birthMonth+"/"+birthYear);
}
}
Answer:
i would say C but i may be wrong have a great day
Explanation:
