A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to
Solution
<p>$B = {\mu _0}ni$</p>
<p>Now $i \to 2i$</p>
<p>And $n \to {n \over 2}$</p>
<p>$B' = {\mu _0}{n \over 2} \times 2i = {\mu _0}ni = B$</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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