JEE Main 2025 (Online) 2nd April Evening Shift | Magnetic Effect of Current Question 9 | Physics

Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current I. Let $\mathrm{B}_2$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x: \mathrm{R}=3: 4, \frac{\mathrm{~B}_2}{\mathrm{~B}_1}$ is :

$64: 125$

$25: 16$

$4: 5$

$16: 25$

JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be :

[a/2, 3a]

[a/4, 3a/2]

[a/2, 2a]

[a/4, 2a]

JEE Main 2025 (Online) 28th January Evening Shift

MCQ (Single Correct Answer)

-1

JEE Main 2025 (Online) 28th January Evening Shift Physics - Magnetic Effect of Current Question 20 English

An infinite wire has a circular bend of radius a, and carrying a current I as shown in the figure. The magnitude of magnetic field at the origin O of the arc is given by:

$\frac{\mu_0}{2 \pi} \frac{\mathrm{I}}{\mathrm{a}}\left[\frac{\pi}{2}+2\right]$

$\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{a}}\left[\frac{3 \pi}{2}+1\right]$

$\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{a}}\left[\frac{\pi}{2}+1\right]$

$\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{a}}\left[\frac{3 \pi}{2}+2\right]$

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