Electromagnetic Induction JEE Main

JEE MAIN 2025

29-01-2025 SHIFT-1

Question

Consider ${{\rm{I}}_1}$ and ${{\rm{I}}_2}$ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If ${{\rm{L}}_1} = $ self inductance of coil $1,{{\rm{M}}_{12}} = $ mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

Select the correct option:

${\varepsilon _1} = - {{\rm{L}}_1}\frac{{{\rm{d}}{{\rm{I}}_2}}}{{{\rm{dt}}}} - {{\rm{M}}_{12}}\frac{{{\rm{d}}{{\rm{I}}_1}}}{{{\rm{dt}}}}$

${\varepsilon _1} = - {{\rm{L}}_1}\frac{{{\rm{d}}{{\rm{I}}_1}}}{{{\rm{dt}}}} + {{\rm{M}}_{12}}\frac{{{\rm{d}}{{\rm{I}}_2}}}{{{\rm{dt}}}}$

${\varepsilon _1} = - {{\rm{L}}_1}\frac{{{\rm{d}}{{\rm{I}}_1}}}{{{\rm{dt}}}} - {{\rm{M}}_{12}}\frac{{{\rm{d}}{{\rm{I}}_1}}}{{{\rm{dt}}}}$

${\varepsilon _1} = - {{\rm{L}}_1}\frac{{{\rm{d}}{{\rm{I}}_1}}}{{{\rm{dt}}}} - {{\rm{M}}_{12}}\frac{{{\rm{d}}{{\rm{I}}_2}}}{{{\rm{dt}}}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

${\phi _1} = {{\rm{L}}_1}{{\rm{I}}_1} + {{\rm{M}}_{12}}{{\rm{I}}_2}$
${\varepsilon _1} = - \frac{{{\rm{d}}{\phi _1}}}{{{\rm{dt}}}} = - {{\rm{L}}_1}\frac{{{\rm{dI}}}}{{{\rm{dt}}}} - {{\rm{M}}_{12}}\frac{{{\rm{d}}{{\rm{I}}_2}}}{{{\rm{dt}}}}$

Question Tags

JEE Main

Physics

Hard

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