JEE Main Magnetic Effect of Current PYQ

11. (JEE Main 2023 (Online) 25th January Evening Shift)

Two long parallel wires carrying currents 8A and 15A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is _____________ $\times 10^{- 6}$ T.

(Given : $\sqrt{2} = 1.4$ )

Correct Answer is 68

JEE Main 2023 (Online) 25th January Evening Shift Physics - Magnetic Effect of Current Question 16 English Explanation

Magnetic fields due to both wires will be perpendicular to each other.

\begin{aligned} B_{1} = \frac{μ_{0} i_{1}}{2 π d} B_{2} = \frac{μ_{0} i_{2}}{2 π d} \\ B_{net} = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{2 π d} \sqrt{i_{1}^{2} + i_{2}^{2}} \\ = \frac{4 π \times 10^{- 7}}{2 π \times (7 / \sqrt{2}) \times 10^{- 2}} \times \sqrt{15^{2} + 8^{2}} (As d = \frac{7}{\sqrt{2}} cm) \\ = 68 \times 10^{- 6} T \end{aligned}

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

A current carrying rectangular loop PQRS is made of uniform wire. The length $P R = Q S = 5 cm$ and $P Q = R S = 100 cm$ . If ammeter current reading changes from I to $2 I$ , the ratio of magnetic forces per unit length on the wire $P Q$ due to wire $R S$ in the two cases respectively $(f_{P Q}^{I} : f_{P Q}^{2 t})$ is:

JEE Main 2023 (Online) 30th January Evening Shift Physics - Magnetic Effect of Current Question 23 English

A. 1 : 4

B. 1 : 3

C. 1 : 2

D. 1 : 5

Correct Answer is Option (A)

Force between two current carrying wire

$= \frac{μ_{0} I_{1} I_{2}}{2 π d} \times L$

Here, $I_{1}$ & $I_{2}$ are equal

$F = \frac{μ_{0} I^{2}}{2 π d} \times L$

$F \propto I^{2}$

$\frac{F_{I}}{F_{2 I}} = \frac{I^{2}}{4 I^{2}} = \frac{1}{4}$

13. (JEE Main 2023 (Online) 24th January Evening Shift)

A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be $\frac{x}{130}$ N. The value of $x$ is ____________.

Correct Answer is 9

JEE Main 2023 (Online) 24th January Evening Shift Physics - Magnetic Effect of Current Question 13 English Explanation

Force on

5 cm

side

= I ℓ B \sin θ

\begin{aligned} = 2 \times \frac{5}{100} \times 0.75 \times \frac{12}{13} \\ = \frac{9}{130} \\ ∴ x = 9 \end{aligned}

14. (JEE Main 2023 (Online) 30th January Morning Shift )

A massless square loop, of wire of resistance $10 Ω$ , supporting a mass of $1 g$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , directed outwards in the shaded region. A dc voltage $V$ is applied to the loop. For what value of $V$ , the magnetic force will exactly balance the weight of the supporting mass of $1 g$ ?

(If sides of the loop $= 10 cm, g = 10 {ms}^{- 2}$ )

JEE Main 2023 (Online) 30th January Morning Shift Physics - Magnetic Effect of Current Question 22 English

A. 1 V

B. $\frac{1}{10}$ V

C. 10V

D. 100V

Correct Answer is Option (C)

For balancing of force

$∴ F_{l o o p} = weight$

$(\frac{V}{R}) I B = m g$

$(\frac{V}{10}) \times \frac{10}{100} \times (10^{3} \times 10^{- 4}) = (\frac{1}{1000}) \times 10$

$V = 10$ volts

15. (JEE Main 2023 (Online) 24th January Morning Shift )

Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F $_{1}$ . If distance between wires is halved and currents on them are doubled, force F $_{2}$ on 10 cm length of wire P will be:

A. $\frac{F_{1}}{8}$

B. 10 F $_{1}$

C. $\frac{F_{1}}{10}$

D. 8 F $_{1}$

Correct Answer is Option (D)

$\begin{aligned} Force per unit length between two parallel straight
wires = \frac{μ_{0} i_{1} i_{2}}{2 π d} \\ \frac{F_{1}}{F_{2}} = \frac{\frac{μ_{0} (10)^{2}}{2 π (5 cm)}}{\frac{μ_{0} (20)^{2}}{2 π (\frac{5 cm}{2})}} = \frac{1}{8} \\ \Rightarrow F_{2} = 8 F_{1} \end{aligned}$