JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 7)
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s-1. At some instant, a part of L is in a uniform
magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 $$\Omega $$, the current in the
loop at that instant will be close to :
_12th_April_Morning_Slot_en_7_1.png)
115 $$\mu $$A
170 $$\mu $$A
60 $$\mu $$A
150 $$\mu $$A
Explanation
Since it is a balanced wheatstone bridge, its equivalent resistance = $${4 \over 3}\Omega $$
$$\varepsilon = B\ell v = 5 \times {10^{ - 4}}V$$
So total resistance
$$R = {4 \over 3} + 1.7 \approx 3\Omega $$
$$ \therefore i = {\varepsilon \over R} \approx 166\,\mu A \approx 170\,\mu A$$
$$\varepsilon = B\ell v = 5 \times {10^{ - 4}}V$$
So total resistance
$$R = {4 \over 3} + 1.7 \approx 3\Omega $$
$$ \therefore i = {\varepsilon \over R} \approx 166\,\mu A \approx 170\,\mu A$$
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