MHT CET · Physics · Magnetic Effects of Current
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
- A \(\sqrt{15} \mathrm{I} \ell \mathrm{B}\)
- B \(\sqrt{11} \mathrm{I} \ell \mathrm{B}\)
- C \(\sqrt{13} \mathrm{I} \ell \mathrm{B}\)
- D \(\sqrt{19} \mathrm{I} \ell \mathrm{B}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{13} \mathrm{I} \ell \mathrm{B}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{F}} =\overrightarrow{\mathrm{I} \ell} \times \overrightarrow{\mathrm{B}}\)
\(\therefore \overrightarrow{\mathrm{F}} =\mathrm{I} \ell \hat{\mathrm{i}} \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{B}\)
\(=\mathrm{I} \ell \mathrm{B}(\hat{\mathrm{i}} \times \hat{\mathrm{i}}+\hat{\mathrm{i}} \times 2 \hat{\mathrm{j}}-3 \hat{\mathrm{i}} \times \hat{\mathrm{k}})=\mathrm{I} \ell \mathrm{B}(0+2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}})\)
\(=(2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}}) \mathrm{I} \ell \mathrm{B}\)
Magnitude of \((2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}})=\sqrt{4+9}=\sqrt{13}\)
\(\therefore \mathrm{F}=\sqrt{13} \mathrm{I} \ell \mathrm{B}\)
\(\therefore \overrightarrow{\mathrm{F}} =\mathrm{I} \ell \hat{\mathrm{i}} \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{B}\)
\(=\mathrm{I} \ell \mathrm{B}(\hat{\mathrm{i}} \times \hat{\mathrm{i}}+\hat{\mathrm{i}} \times 2 \hat{\mathrm{j}}-3 \hat{\mathrm{i}} \times \hat{\mathrm{k}})=\mathrm{I} \ell \mathrm{B}(0+2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}})\)
\(=(2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}}) \mathrm{I} \ell \mathrm{B}\)
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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