JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 11)
A rigid square loop of side 'a' and carrying
current I2 is lying on a horizontal surface near
a long current I1 carrying wire in the same plane
as shown in figure. The net force on the loop
due to wire will be :
_9th_April_Morning_Slot_en_11_1.png)
Repulsive and equal to $$\mu $$0I1I2/4$$\pi $$
Attractive and equal to $$\mu $$0I1I2/3$$\pi $$
Repulsive and equal to $$\mu $$0I1I2/2$$\pi $$
Zero
Explanation
F3 & F4 cancel each other.
Force on PQ will be F1 = I2B1 a
= $${I_2}{{{\mu _0}{I_1}} \over {2\pi a}}a$$
= $${{{\mu _0}{I_1}} \over {2\pi a}}a = {{{\mu _0}{I_1}{I_2}} \over {2\pi }}$$
Force on RS will be F2 = I2 B2 a
= $${I_2}{{{\mu _0}{I_1}} \over {2\pi 2a}}a = {{{\mu _0}{I_1}{I_2}} \over {4\pi }}$$
Net force = F1 – F2 = $${{{\mu _0}{I_1}{I_2}} \over {4\pi }}$$ repulsion
Force on PQ will be F1 = I2B1 a
= $${I_2}{{{\mu _0}{I_1}} \over {2\pi a}}a$$
= $${{{\mu _0}{I_1}} \over {2\pi a}}a = {{{\mu _0}{I_1}{I_2}} \over {2\pi }}$$
Force on RS will be F2 = I2 B2 a
= $${I_2}{{{\mu _0}{I_1}} \over {2\pi 2a}}a = {{{\mu _0}{I_1}{I_2}} \over {4\pi }}$$
Net force = F1 – F2 = $${{{\mu _0}{I_1}{I_2}} \over {4\pi }}$$ repulsion
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