JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 12)
A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of
a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field
changes with time at a steady rate $${{dB} \over {dt}}$$ = 0.032 Ts–1. The induced current in the loop is close to
(Resistivity of the metal wire is 1.23 $$ \times $$ 10–8 $$\Omega $$m)
0.53 A
0.43 A
0.34 A
0.61 A
Explanation
We know, $$\phi = BA$$
Also, $$E = {{d\phi } \over {dt}} = {{AdB} \over {dt}}$$
$$E = {l^2}{{dB} \over {dt}}$$
$$i = {E \over R} $$
= $${{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}$$
$$= {{{l^2}} \over {pl}}{{dB} \over {dt}}A$$
= $${{{{l^2}\pi {R^2}} \over {\rho l}}{{dB} \over {dt}}}$$
$$ \therefore $$ $$i = {{30} \over 4} \times {{30} \over 4} \times {{{{10}^{ - 4}} \times 0.032 \times 4 \times {{10}^{ - 6}} \times \pi } \over {1.23 \times {{10}^{ - 8}} \times 30 \times {{10}^{ - 2}} \times {{10}^3}}}$$
$$ \Rightarrow $$ $$i = {{240 \times \pi \times {{10}^{ - 10}}} \over {1.23 \times {{10}^{ - 7}}}}$$
$$ \Rightarrow $$ $$i = {{240 \times 3.14 \times {{10}^{ - 3}}} \over {1.23}}$$
$$ = {{753.6} \over {1.23}} \times {10^{ - 3}}$$
$$ \Rightarrow $$ $$i = 612.68 \times {10^{ - 3}} = 0.61A$$
Also, $$E = {{d\phi } \over {dt}} = {{AdB} \over {dt}}$$
$$E = {l^2}{{dB} \over {dt}}$$
$$i = {E \over R} $$
= $${{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}$$
$$= {{{l^2}} \over {pl}}{{dB} \over {dt}}A$$
= $${{{{l^2}\pi {R^2}} \over {\rho l}}{{dB} \over {dt}}}$$
$$ \therefore $$ $$i = {{30} \over 4} \times {{30} \over 4} \times {{{{10}^{ - 4}} \times 0.032 \times 4 \times {{10}^{ - 6}} \times \pi } \over {1.23 \times {{10}^{ - 8}} \times 30 \times {{10}^{ - 2}} \times {{10}^3}}}$$
$$ \Rightarrow $$ $$i = {{240 \times \pi \times {{10}^{ - 10}}} \over {1.23 \times {{10}^{ - 7}}}}$$
$$ \Rightarrow $$ $$i = {{240 \times 3.14 \times {{10}^{ - 3}}} \over {1.23}}$$
$$ = {{753.6} \over {1.23}} \times {10^{ - 3}}$$
$$ \Rightarrow $$ $$i = 612.68 \times {10^{ - 3}} = 0.61A$$
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