Source And Effects Of Magnetic Field | Physics JEE previous year questions with Solutions

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Topic Question Set

Q 11 :

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 [2023]

[IMAGE 226]

(Neglect the effect of earth's magnetic field.)

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

1

2

3

4

(4)

$d \tan 60 ° = 2 \sqrt{3}$

$d = 2 cm$

$B = 3 \times \frac{μ_{0} i}{2 π d} \sin 60 °$

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

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

Q 12 :

Two long parallel wires carrying currents 8 A and 15 A 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: $\sqrt{2} = 1.4$ ) [2023]

0%

(68)

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

[IMAGE 227]

$B_{1} = \frac{μ_{0} i_{1}}{2 π d}, B_{2} = \frac{μ_{0} i_{2}}{2 π d}$

$B_{net} = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{2 π d} \sqrt{i_{1}^{2} + i_{2}^{2}}$

$\Rightarrow \frac{4 π \times 10^{- 7}}{2 π \times (\frac{7}{\sqrt{2}}) \times 10^{- 2}} \times \sqrt{8^{2} + 15^{2}} (d = \frac{7}{\sqrt{2}} cm)$

$= 68 \times 10^{- 6} T$

Q 13 :

An electron in a hydrogen atom revolves around its nucleus with a speed of $6.76 \times 10^{6} {ms}^{- 1}$ in an orbit of radius $0.52 Å$ . The magnetic field produced at the nucleus of the hydrogen atom is _________ T. [2023]

0%

(40)

$Magnetic field due to moving charge$

$B = \frac{μ_{0}}{4 π} \frac{q v \sin θ}{r^{2}}$

$B = \frac{μ_{0}}{4 π} \frac{e v \sin (π / 2)}{r^{2}}$

$B = \frac{10^{- 7} \times 1.6 \times 10^{- 19} \times 6.76 \times 10^{6}}{0.52 \times 0.52 \times 10^{- 20}}$

$\Rightarrow B = 40 T$

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