Question

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

Short answer

The peak magnetic flux through the solenoid is$1.72\times {10}^{-7}Wb$.

Step-by-step solution

Given Information

We are given that $110\Omega$resistor ,$5.0-cm-$long solenoid with $1.0-cm-$diameter . The no of turns are given as $800$. We are given that the source has$12V$ voltage andlocalid="1650110781883" $5.0kHz$localid="1650111296897" $5kHz$frequency .

We need to find peak magnetic flux .

Simplify

Firstly , we will find area of cross-section $\Rightarrow A=\mathrm{\pi }\times r^2=\mathrm{\pi }\times {\left(0.5\times {10}^{-2}\mathrm{m}\right)}^2=0.78\times {10}^{-4}{\mathrm{m}}^2$Then , we will calculate current , $R=\frac{{\epsilon }_0}{I}\Rightarrow I=\frac{{\epsilon }_0}{R}=\frac{12}{110}=0.108A$ .

now we will calculate magnetic field ,$B=\mu nI$ ,here $n$$n$is no of turns per unit length, $I$$I$is current.

localid="1650111307310" $B=4\mathrm{\pi }\times {10}^{-7}\times \left(\frac{800}{0.05}\right)\times 0.109=0.00219\mathrm{T}$

also magnetic flux ,$\varphi =BA=0.00219\times 0.78\times {10}^{-4}Wb=1.72\times {10}^{-7}Wb$.