Question

A 10-turn ideal solenoid has an inductance of 3.5 mH. When the solenoid carries a current of 2.0 A the magnetic flux through each turn is:

A) 7.0 x 10 2 Wb
B) 0 Wb
C) 3.5 x 104 wb
D) 70 x 10-3w
E) 7.0 x 104Wb

Answer 1

The magnetic flux through each turn of the 10-turn ideal solenoid with 3.5 mH inductance and a 2.0 A current is 7.0 x 10^-4 Weber, which corresponds to option A.

Step-by-step explanation:

The magnetic flux through each turn of an ideal solenoid can be calculated using the formula for magnetic flux Φ = LI/n, where L is inductance, I is current, and n is the number of turns. Given that the inductance L is 3.5 mH (or 3.5 x 10-3 H), the current I is 2.0 A, and the solenoid has n = 10 turns, we can plug these values into the equation to find the magnetic flux Φ.

Therefore, the magnetic flux through one turn is:

Φ = (3.5 x 10

-3

H)(2.0 A) / 10 = 7.0 x 10

-4

Wb

Which matches option A when you adjust the exponent to common scientific notation, thus: 7.0 x 102 x 10-6 Wb or 7.0 x 10-4 Wb.