JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 22)
As shown in the figure, two infinitely long, identical wires are bent by 90o and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4cm, and the magnitude of the magnetic field at O is 10–4 T, and the two wires carry equal
currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be ($$\mu $$0 = 4$$\pi $$ $$ \times $$ 10–7 NA–2) :
_12th_January_Morning_Slot_en_22_1.png)
40 A, perpendicular into the page
40 A, perpendicular out of the page
20 A, perpendicular into the page
40 A, perpendicular out of the page
설명
Magnetic field at 'O' will be done to 'PS' and 'QN' Only
i.e. B0 = BPS + BQN $$ \to $$ Both inwards
Let current in each wire = i
$$ \therefore $$ B0 = $${{{\mu _0}i} \over {4\pi d}} + {{{\mu _0}i} \over {4\pi d}}$$
or 10$$-$$4 = $${{{\mu _0}i} \over {2\pi d}}$$ = $${{2 \times {{10}^{ - 7}} \times i} \over {4 \times {{10}^{ - 2}}}}$$
$$ \therefore $$ i = 20 A
i.e. B0 = BPS + BQN $$ \to $$ Both inwards
Let current in each wire = i
$$ \therefore $$ B0 = $${{{\mu _0}i} \over {4\pi d}} + {{{\mu _0}i} \over {4\pi d}}$$
or 10$$-$$4 = $${{{\mu _0}i} \over {2\pi d}}$$ = $${{2 \times {{10}^{ - 7}} \times i} \over {4 \times {{10}^{ - 2}}}}$$
$$ \therefore $$ i = 20 A