JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 13)
A conducting circular loop made of a thin wire, has area 3.5 $$ \times $$ 10$$-$$3 m2 and resistance 10 $$\Omega $$. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T)sin(50$$\pi $$t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to :
0.14 mC
0.7 mC
0.21 mC
0.6 mC
Explanation
At t = 0 s
B(0) = 0.4 sin (0) = 0
and at t = 10 ms
B(10) = 0.4 sin (50$$\pi $$$$ \times $$10$$ \times $$10-3)
= 0.4 sin $$\left( {{\pi \over 2}} \right)$$
= 0.4
As q = $${{\Delta \phi } \over R}$$
= $${{A\left[ {B\left( {10} \right) - B\left( 0 \right)} \right]} \over {10}}$$
= $${{3.5 \times {{10}^{ - 3}}\left[ {0.4 - 0} \right]} \over {10}}$$
= 0.14 mC
B(0) = 0.4 sin (0) = 0
and at t = 10 ms
B(10) = 0.4 sin (50$$\pi $$$$ \times $$10$$ \times $$10-3)
= 0.4 sin $$\left( {{\pi \over 2}} \right)$$
= 0.4
As q = $${{\Delta \phi } \over R}$$
= $${{A\left[ {B\left( {10} \right) - B\left( 0 \right)} \right]} \over {10}}$$
= $${{3.5 \times {{10}^{ - 3}}\left[ {0.4 - 0} \right]} \over {10}}$$
= 0.14 mC
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