In given figure, a wire loop has been bent so that it has three segments $\mathrm{a}\mathrm{b}$ (a quarter circle), $\mathrm{b}\mathrm{c}$ (a square corner) & $\mathrm{c}\mathrm{a}$ (straight line). Here are three choices for a magnetic field through the loop -



(1) ${\overrightarrow{\mathrm{B}}}_1=3\widehat{\mathrm{i}}\mathrm{ }+\mathrm{ }7\widehat{\mathrm{j}}\mathrm{ }–\mathrm{ }5\mathrm{t}\widehat{\mathrm{k}}$

(2) ${\overrightarrow{\mathrm{B}}}_2=5\mathrm{t}\widehat{\mathrm{i}}\mathrm{ }–\mathrm{ }4\widehat{\mathrm{j}}\mathrm{ }–\mathrm{ }15\widehat{\mathrm{k}}$

(3) ${\overrightarrow{\mathrm{B}}}_3=2\widehat{\mathrm{i}}\mathrm{ }–5\mathrm{t}\widehat{\mathrm{j}}\mathrm{ }–\mathrm{ }12\widehat{\mathrm{k}}$

Where $\mathrm{B}$ is in milli tesla and $\mathrm{t}$ is in second. If the induced current in the loop due to $\overrightarrow{{\mathrm{B}}_1},\overrightarrow{{\mathrm{B}}_2}\overrightarrow{{,\mathrm{B}}_3}$ are ${\mathrm{i}}_1,\mathrm{ }{\mathrm{i}}_2,\mathrm{ }{\mathrm{i}}_3$ respectively then -

• A

  ${\mathrm{i}}_1>\mathrm{ }{\mathrm{i}}_2>\mathrm{ }{\mathrm{i}}_3$
• Correct answer
  B

  ${\mathrm{i}}_2>\mathrm{ }{\mathrm{i}}_1>\mathrm{ }{\mathrm{i}}_3$
• C

  ${\mathrm{i}}_3\mathrm{ }>\mathrm{ }{\mathrm{i}}_2\mathrm{ }>\mathrm{ }{\mathrm{i}}_1$
• D

  ${\mathrm{i}}_1={\mathrm{i}}_2={\mathrm{i}}_3$

Hence, the correct answer is option: 'B'.