The magnetic field inside a 200 turns solenoid of radius 10 cm is $2.9 \times 10^{-4} ~\mathrm{Tesla}$. If the solenoid carries a current of 0.29 A , then the length of the solenoid is _______ $\pi \mathrm{cm}$.
Answer (integer)
8
Solution
<p>We know, magnetic field due to solenoid, $B = {\mu _0}nI$</p>
<p>where, I = current, $n = {N \over L}$ = no. of turns per unit length</p>
<p>$\Rightarrow B = {{{\mu _0}NI} \over L}$</p>
<p>$$ \Rightarrow L = {{{\mu _0}NI} \over B} = {{4\pi \times {{10}^{ - 7}} \times 200 \times 0.29} \over {2.9 \times {{10}^{ - 4}}}}$$</p>
<p>$\Rightarrow L = 8\pi \times {10^{ - 2}}\,m$</p>
<p>$\Rightarrow L = 8\pi \,cm$</p>
<p>Hence, answer = 8.</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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