A straight wire carrying current 10 A is lying such that position vector at its ends are (r1¯=2i^−2j^) m and (r2¯=10i^+4j^) m. It is kept in a uniform magnetic field 1Tk^ Then the force acting on wire is

l→=r2→−r1→=8i^+6j^F→=i(l→×B→)

A straight wire carrying current 10 A is lying such that position vector at its ends are (r1¯=2i^−2j^) m and (r2¯=10i^+4j^) m. It is kept in a uniform magnetic field 1Tk^ Then the force acting on wire is

l→=r2→−r1→=8i^+6j^F→=i(l→×B→)

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A straight wire carrying current 10 A is lying such that position vector at its ends are $(\bar{r_{1}} = 2 \hat{i} - 2 \hat{j}) m$ and $(\bar{r_{2}} = 10 \hat{i} + 4 \hat{j}) m$ . It is kept in a uniform magnetic field $1 T \hat{k}$ Then the force acting on wire is

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$\vec{l} = \vec{r_{2}} - \vec{r_{1}} = 8 \hat{i} + 6 \hat{j}$

$\vec{F} = i (\vec{l} \times \vec{B})$

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A straight wire carrying current 10 A is lying such that position vector at its ends are (r1¯=2i^−2j^) m and (r2¯=10i^+4j^) m. It is kept in a uniform magnetic field 1Tk^ Then the force acting on wire is

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A straight wire carrying current 10 A is lying such that position vector at its ends are $(\bar{r_{1}} = 2 \hat{i} - 2 \hat{j}) m$ and $(\bar{r_{2}} = 10 \hat{i} + 4 \hat{j}) m$ . It is kept in a uniform magnetic field $1 T \hat{k}$ Then the force acting on wire is