A conducting wire bent in the form of a parab | vedclass

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

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$ 32\,\hat{i}$

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(B) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the vector displacement from the starting point to the ending point of the wire.From the figure,the wire starts at point $A(2, 2)$ and ends at point $B(2, -2)$.Therefore,the effective length vector is $\vec{L}_{eff} = \vec{r}_B - \vec{r}_A = (2\hat{i} - 2\hat{j}) - (2\hat{i} + 2\hat{j}) = -4\hat{j} \, m$.The current $i = 2 \, A$ and the magnetic field $\vec{B} = -4\hat{k} \, T$.Substituting these values into the force formula:$\vec{F} = 2 \, A 	imes (-4\hat{j} \, m) 	imes (-4\hat{k} \, T)$$\vec{F} = 2 	imes (-4) 	imes (-4) 	imes (\hat{j} 	imes \hat{k})$Since $\hat{j} 	imes \hat{k} = \hat{i}$,we get:$\vec{F} = 32\hat{i} \, N$.

$A$ conducting wire bent in the form of a parabola $y^2 = 2x$ carries a current $i = 2 \, A$ as shown in the figure. This wire is placed in a uniform magnetic field $\vec{B} = -4\,\hat{k} \, T$. The magnetic force on the wire is (in newton):

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