A current-carrying wire LN is bent as shown b | vedclass

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(B) The force on a current-carrying wire in a uniform magnetic field is given by $F = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the ef

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$5$

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(B) The force on a current-carrying wire in a uniform magnetic field is given by $F = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the effective length vector from the starting point $L$ to the end point $N$.From the geometry,the horizontal displacement is $6 \, cm$ and the total vertical displacement is $4 \, cm + 4 \, cm = 8 \, cm$.The effective length $L_{eff} = \sqrt{(6 \, cm)^2 + (8 \, cm)^2} = \sqrt{36 + 64} \, cm = \sqrt{100} \, cm = 10 \, cm = 0.1 \, m$.The force magnitude is $F = B I L_{eff} = 5 \, T 	imes 10 \, A 	imes 0.1 \, m = 5 \, N$.

$A$ current-carrying wire $LN$ is bent as shown below. If the wire carries a current of $10 \, A$ and is placed in a magnetic field of $5 \, T$ which acts perpendicular to the paper outwards,then it will experience a force of ......... $N$.

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