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  • A current carrying loop in the form of a right-angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F what is the force on the

A current carrying loop in the form of a right-angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F what is the force on the arm AC?

Option: 1

-\sqrt{2} \vec{F}


Option: 2

-\vec{F}


Option: 3

\vec{F}


Option: 4

\sqrt{2} \vec{F}


Answers (1)

best_answer

Let a current i be flowing in the loop ABC in the direction shown in the figure. If the length of each of the sides AB and BC be x then |\vec{F}|=i \times B
where B is the magnitude of the magnetic force.
The direction of \vec{F} will be in the direction perpendicular to the plane of the paper and going into it.
By Pythagoras' theorem,
\mathrm{AC}=\sqrt{\mathrm{x}^2+\mathrm{x}^2}=\sqrt{2} \mathrm{x}
\therefore Magnitude of force on AC
\begin{gathered} =\mathrm{i} \sqrt{2} \times B \sin 45^{\circ} \\ =\mathrm{i} \sqrt{2} \times \mathrm{B} \times \frac{1}{\sqrt{2}} \\ =\mathrm{ixB}=|\overrightarrow{\mathrm{F}}| \end{gathered}
The direction of the force on AC is perpendicular to the plane of the paper and going out of it. Hence, force on \mathrm{AC}=-\overrightarrow{\mathrm{F}}

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SANGALDEEP SINGH

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