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  • A current carrying closed 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 i

A current carrying closed 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

\begin{aligned} & \text { Here, } \vec{F}_{B C}=\vec{F} \\ & \because \quad \vec{F}_{A B}=0 \end{aligned}

The net magnetic force on a current carrying a closed loop in a uniform magnetic field is zero. 

\begin{aligned} & \therefore \vec{F}_{A B}+\vec{F}_{B C}+\vec{F}_{A C}=0 \\ & \Rightarrow \vec{F}_{A C}=-\vec{F}_{B C} \quad\left(\because \vec{F}_{A B}=0\right) \end{aligned}

 

Posted by

Ritika Jonwal

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