Question:
As shown in the figure, two infinitely long, identical wires are bent by \(90^\circ\) 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 = 4 cm, 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 (\(\mu_0 = 4\pi \times 10^{-7}\ \mathrm{NA^{-2}}\)) :
As shown in the figure, two infinitely long, identical wires are bent by \(90^\circ\) 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 = 4 cm, 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 (\(\mu_0 = 4\pi \times 10^{-7}\ \mathrm{NA^{-2}}\)) :
Show Hint
O lies on the line of the straight segments LP and QM, so they contribute nothing; each wire contributes only through its semi-infinite vertical arm at distance 4 cm, with \(B=\mu_0 I/4\pi d\). Use the right-hand rule for direction.
Updated On: Nov 5, 2026
- 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
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The Correct Option is C
Solution and Explanation
Segments LP and QM are along the x-axis and O is on the x-axis, so for them \(d\vec{l}\times\vec{r}=0\): no field at O.
Arm PS is a semi-infinite wire ending at P (the foot of the perpendicular from O) at perpendicular distance \(d = OP = 4\ \mathrm{cm} = 0.04\) m. Its field at O: \(B_1 = \dfrac{\mu_0 I}{4\pi d}\). Same for arm QN: \(B_2 = \dfrac{\mu_0 I}{4\pi d}\).
Direction: in PS the current is along +y and O is to the right (+x) of it: \(\hat{y}\times\hat{x} = -\hat{z}\), i.e. into the page. In QN the current is along \(-y\) and O is to the left (\(-x\)) of it: \((-\hat{y})\times(-\hat{x}) = -\hat{z}\), again into the page. So the two fields add.
\(B = 2\cdot\dfrac{\mu_0 I}{4\pi d} = 2\times 10^{-7}\times\dfrac{I}{0.04} = 5\times 10^{-6}\,I\).
Given \(B = 10^{-4}\) T: \(I = \dfrac{10^{-4}}{5\times 10^{-6}} = 20\) A, field directed into the page.
Answer: (3) 20 A, perpendicular into the page
Arm PS is a semi-infinite wire ending at P (the foot of the perpendicular from O) at perpendicular distance \(d = OP = 4\ \mathrm{cm} = 0.04\) m. Its field at O: \(B_1 = \dfrac{\mu_0 I}{4\pi d}\). Same for arm QN: \(B_2 = \dfrac{\mu_0 I}{4\pi d}\).
Direction: in PS the current is along +y and O is to the right (+x) of it: \(\hat{y}\times\hat{x} = -\hat{z}\), i.e. into the page. In QN the current is along \(-y\) and O is to the left (\(-x\)) of it: \((-\hat{y})\times(-\hat{x}) = -\hat{z}\), again into the page. So the two fields add.
\(B = 2\cdot\dfrac{\mu_0 I}{4\pi d} = 2\times 10^{-7}\times\dfrac{I}{0.04} = 5\times 10^{-6}\,I\).
Given \(B = 10^{-4}\) T: \(I = \dfrac{10^{-4}}{5\times 10^{-6}} = 20\) A, field directed into the page.
Answer: (3) 20 A, perpendicular into the page
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