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A wire carrying current 'I' along $\mathrm{x}$ axis has length ' $\ell$ ' and it is kept in a magnetic
field $\vec{B}=(\hat{\imath}+2 \hat{\jmath}-3 \hat{k}) B \frac{W b}{m^{2}}$. The magnitude of magnetic force acting on the wire
is
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field $\vec{B}=(\hat{\imath}+2 \hat{\jmath}-3 \hat{k}) B \frac{W b}{m^{2}}$. The magnitude of magnetic force acting on the wire
is
Solution:
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Verified Answer
The correct answer is:
$\sqrt{13} \mathrm{I} \ell \mathrm{B}$
$\begin{aligned} \overrightarrow{\mathrm{F}} &=\overrightarrow{\mathrm{I} \ell} \times \overrightarrow{\mathrm{B}} \\ \therefore \quad \overrightarrow{\mathrm{F}} &=\mathrm{I} \ell \hat{\mathrm{i}} \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{B} \\ &=\mathrm{I} \ell \mathrm{B}(\hat{\mathrm{i}} \times \hat{\mathrm{i}}+\hat{\mathrm{i}} \times 2 \hat{\mathrm{j}}-3 \hat{\mathrm{i}} \times \hat{\mathrm{k}})=\mathrm{I} \ell \mathrm{B}(0+2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}}) \\ &=(2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}}) \mathrm{I} \ell \mathrm{B} \end{aligned}$
Magnitude of $(2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}})=\sqrt{4+9}=\sqrt{13}$
$\therefore \mathrm{F}=\sqrt{13} \mathrm{I} \ell \mathrm{B}$
Magnitude of $(2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}})=\sqrt{4+9}=\sqrt{13}$
$\therefore \mathrm{F}=\sqrt{13} \mathrm{I} \ell \mathrm{B}$
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