A long wire carrying current $\sqrt[]{2}$ A is placed in uniform magnetic field of 3 × 10^–5 T. If magnetic field is perpendicular to wire, find the magnetic force on unit length of wire.

Correct Option - 2


Detailed Solution:

F = iL B sinq

= $\sqrt[]{2}$  × 1 × 3 × 10^–5 sin90

= 3 $\sqrt[]{2}$  × 10^–5  N

<div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1758869041phpritG2E.jpeg" alt="" width="139" height="94" loading="lazy"></picture></div></div><p><span lang="EN-IN">F = iL B sin</span><span lang="EN-IN">q</span></p><p><span lang="EN-IN">= <span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span>&nbsp;× 1 × 3 × 10^–5 sin90</span></p><p><span lang="EN-IN">= 3<span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span>&nbsp;× 10^–5 &nbsp;N</span></p>

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