JEE Main Magnetic Effect of Current PYQ

6. (JEE Main 2024 (Online) 30th January Evening Shift )

The current of $5 A$ flows in a square loop of sides $1 m$ is placed in air. The magnetic field at the centre of the loop is $X \sqrt{2} \times 10^{- 7} T$ . The value of $X$ is _________.

Correct answer is 40

$\begin{aligned} B = 4 \times \frac{μ_{0} i}{4 π (1 / 2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) \\ = 4 \times 10^{- 7} \times 5 \times 2 \times \sqrt{2} \\ - 40 \sqrt{2} \times 10^{- 7} T \end{aligned}$

7. (JEE Main 2024 (Online) 27th January Evening Shift )

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_{1} = 2 π m$ and $R_{2} = 4 π m$ , carrying current $I = 4 A$ as per figure given below is $α \times 10^{- 7} T$ . The value of $α$ is ________. (Centre $O$ is common for all segments)

JEE Main 2024 (Online) 27th January Evening Shift Physics - Magnetic Effect of Current Question 27 English

Correct answer is 3

JEE Main 2024 (Online) 27th January Evening Shift Physics - Magnetic Effect of Current Question 27 English Explanation

$\begin{aligned} \frac{μ_{0} i}{2 R_{2}} (\frac{π}{2 π}) \otimes + \frac{μ_{0} i}{2 R_{1}} (\frac{π}{2 π}) \otimes \\ (\frac{μ_{0} i}{4 R_{2}} + \frac{μ_{0} i}{4 R_{1}}) \otimes \\ \frac{4 π \times 10^{- 7} \times 4}{4 \times 4 π} + \frac{4 π \times 10^{- 7} \times 4}{4 \times 2 π} \\ = 3 \times 10^{- 7} = α \times 10^{- 7} \\ α = 3 \end{aligned}$

8. (JEE Main 2024 (Online) 27th January Morning Shift )

Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is $5.0 cm$ . The magnitude of the magnetic field at a point $P$ midway between the wires is _______ $μ T$

(Given : $μ_{0} = 4 π \times 10^{- 7} {TmA}^{- 1}$ )

JEE Main 2024 (Online) 27th January Morning Shift Physics - Magnetic Effect of Current Question 28 English

Correct answer is 160

$\begin{aligned} B = (\frac{μ_{0} i}{2 π a}) \times 2 = \frac{4 π \times 10^{- 7} \times 10}{π \times (\frac{5}{2} \times 10^{- 2})} \\ = 16 \times 10^{- 5} = 160 μ T \end{aligned}$

9. (JEE Main 2023 (Online) 10th April Evening Shift)

A straight wire carrying a current of $14 A$ is bent into a semi-circular arc of radius $2.2 cm$ as shown in the figure. The magnetic field produced by the current at the centre $(O)$ of the arc. is ____________ $\times 10^{- 4} T$

JEE Main 2023 (Online) 10th April Evening Shift Physics - Magnetic Effect of Current Question 6 English

Correct Answer 2

$\begin{aligned} B_{at O} = \frac{μ_{0} I}{4 R} = \frac{4 π \times 10^{- 7} \times 14}{4 \times 2.2 \times 10^{- 2}} \\ = 2 \times 10^{- 4} T \end{aligned}$

10. (JEE Main 2023 (Online) 8th April Evening Shift)

The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of coil on its axis is $\sqrt{x} : 1$ . The value of $x$ is __________

Correct Answer is 8

The magnetic field at the center of a loop (B₁) is given by

$B_{1} = \frac{μ_{0} I}{2 r}$

The magnetic field on the axis of the loop at a distance ( r ) from the center (B₂) is given by

$B_{2} = \frac{μ_{0} I r^{2}}{2 (r^{2} + d^{2})^{3 / 2}}$

where ( d ) is the distance from the center of the coil along the axis. Since ( d = r ), we get

$B_{2} = \frac{μ_{0} I}{4 \sqrt{2} r}$

The ratio of $B_{1}$ to $B_{2}$ is

$\frac{B_{1}}{B_{2}} = \frac{μ_{0} I}{2 r} \times \frac{4 \sqrt{2} r}{μ_{0} I} = \sqrt{8} : 1$

So, the value of ( x ) is 8.