JEE Main Magnetic Effect of Current PYQ

11. (JEE Main 2023 (Online) 6th April Morning Shift)

Two identical circular wires of radius $20 cm$ and carrying current $\sqrt{2} A$ are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is __________ $\times 10^{- 8} T$ .

JEE Main 2023 (Online) 6th April Morning Shift Physics - Magnetic Effect of Current Question 3 English

(Take $π = 3.14$ )

Correct Answer is 628

Currently no explanation available

12. (JEE Main 2023 (Online) 1st February Evening Shift )

As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{π}{10}$ m is carrying current $I = 3 A$ . The magnitude of the magnetic field, at the center O of the arc is :

(The permeability of the vacuum $= 4 π \times 10^{- 7} {NA}^{- 2}$ )

JEE Main 2023 (Online) 1st February Evening Shift Physics - Magnetic Effect of Current Question 29 English

A. $4 μ T$

B. $3 μ T$

C. $6 μ T$

D. $1 μ T$

Correct Answer is Option ()

13. (JEE Main 2023 (Online) 1st February Morning Shift )

Find the magnetic field at the point $P$ in figure. The curved portion is a semicircle connected to two long straight wires.

JEE Main 2023 (Online) 1st February Morning Shift Physics - Magnetic Effect of Current Question 26 English

A. $\frac{μ_{0} i}{2 r} (\frac{1}{2} + \frac{1}{2 π})$

B. $\frac{μ_{o} i}{2 r} (1 + \frac{2}{π})$

C. $\frac{μ_{0} \dot{i}}{2 r} (\frac{1}{2} + \frac{1}{π})$

D. $\frac{μ_{0} i}{2 r} (1 + \frac{1}{π})$

Correct Answer is Option (A)

$B_{P} = (\frac{μ_{0} i}{4 r} + \frac{μ_{0} i}{4 π r}) = \frac{μ_{0} i}{2 r} (\frac{1}{2} + \frac{1}{2 π})$

14. (JEE Main 2023 (Online) 31st January Evening Shift )

A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

A. $N^{2} : n^{2}$

B. $N : n$

C. $n : N$

D. $n^{2} : N^{2}$

Correct Answer is Option (A)

$I = (2 π r) n$

$\begin{aligned} r \propto (\frac{I}{n}) \\ B = n (\frac{μ_{0} i}{2 r}) \propto (\frac{μ_{0} i}{2 L}) n^{2} \\ \frac{B_{1}}{B_{2}} = (\frac{N^{2}}{n^{2}}) \end{aligned}$

15. (JEE Main 2023 (Online) 30th January Evening Shift )

As shown in the figure, a current of $2 A$ flowing in an equilateral triangle of side $4 \sqrt{3} cm$ . The magnetic field at the centroid $O$ of the triangle is

JEE Main 2023 (Online) 30th January Evening Shift Physics - Magnetic Effect of Current Question 24 English

(Neglect the effect of earth's magnetic field)

A. $4 \sqrt{3} \times 10^{- 4} T$

B. $4 \sqrt{3} \times 10^{- 5} T$

C. $3 \sqrt{3} \times 10^{- 5} T$

D. $\sqrt{3} \times 10^{- 4} T$

Correct Answer is Option (C)

$B_{n e t} = \frac{μ_{0} i}{4 π r} (\sin α + \sin β) \times 3$

$= \frac{μ_{0} \times 2}{4 π \times (2 \times 10^{- 2})} \times (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) \times 3$

$= 10^{- 7} \times 10^{2} (3 \sqrt{3})$

$= 3 \sqrt{3} \times 10^{- 5}$ T