As shown in the figure, a metallic rod of linear density 0.45 km^-1 is lying horizontally on a smooth incline plane which makes an angle of 45° 5with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be:

{Use g = m/s²}

<p>Linear density, <span contenteditable="false"> <math> <mrow> <mi>λ</mi> <mo>=</mo> </mrow> </math> </span> <math><mrow></mrow></math>&nbsp;0.45 kg/m</p><p><img ></p><p>Let length = <span contenteditable="false"> <math> <mrow> <mi mathvariant="script">l</mi> </mrow> </math> </span><span contenteditable="false"> <math> <mrow> <mo>∴</mo> </mrow> </math> </span><math><mrow></mrow></math>&nbsp;<math><mrow></mrow></math>&nbsp;= m = 0.45 <span contenteditable="false"> <math> <mrow> <mi mathvariant="script">l</mi> </mrow> </math> </span><math><mrow></mrow></math></p><p>B = 0.15 T</p><p>For equilibrium of rod :-</p><p>mg sin 45° = FB sin 45°</p><p><span contenteditable="false"> <math> <mrow> <mo>⇒</mo> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>.</mo> <mn>4</mn> <mn>5</mn> <mi mathvariant="script">l</mi> </mrow> <mo>)</mo> </mrow> <mi>g</mi> <mo>=</mo> <mi>l</mi> <mi mathvariant="script">l</mi> <mi>B</mi> </mrow> </math> </span></p><p>So, <span contenteditable="false"> <math> <mrow> <mi>l</mi> <mo>=</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>4</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mn>0</mn> <mi>A</mi> </mrow> </math> </span><math><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow></math></p>

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