A series RLcircuit is built with a role="math" localid="1650111235060" 110resistor and a 5.0-cm- long,1.0-cm-diameter solenoid with 800turns of wire. What is the peak magnetic flux through the solenoid if the circuit is driven by a 12V,5.0kHzsource?

Short Answer

Expert verified

The peak magnetic flux through the solenoid is1.72×10-7Wb.

Step by step solution

01

Given Information

We are given that 110resistor ,5.0-cm-long solenoid with 1.0-cm-diameter . The no of turns are given as 800. We are given that the source has12V voltage andlocalid="1650110781883" 5.0kHzlocalid="1650111296897" 5kHzfrequency .

We need to find peak magnetic flux .

02

Simplify

Firstly , we will find area of cross-section A=π×r2=π×0.5×10-2m2=0.78×10-4m2Then , we will calculate current , R=ε0II=ε0R=12110=0.108A .

now we will calculate magnetic field ,B=μnI ,here nnis no of turns per unit length, IIis current.

localid="1650111307310" B=4π×10-7×8000.05×0.109=0.00219T

also magnetic flux ,ϕ=BA=0.00219×0.78×10-4Wb=1.72×10-7Wb.

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