JEE Main 2024 (Online) 8th April Evening Shift | Magnetic Effect of Current Question 5 | Physics

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $$\frac{a}{2}$$ and $$2 a$$ from axis of the wire is :

$$4: 1$$

$$3: 4$$

$$1: 1$$

$$1: 4$$

JEE Main 2024 (Online) 6th April Morning Shift

MCQ (Single Correct Answer)

-1

An element $$\Delta l=\Delta x\hat{i}$$ is placed at the origin and carries a large current $$I=10 \mathrm{~A}$$. The magnetic field on the $$y$$-axis at a distance of $$0.5 \mathrm{~m}$$ from the elements $$\Delta x$$ of $$1 \mathrm{~cm}$$ length is:

JEE Main 2024 (Online) 6th April Morning Shift Physics - Magnetic Effect of Current Question 1 English

$$10 \times 10^{-8} \mathrm{~T}$$

$$8 \times 10^{-8} \mathrm{~T}$$

$$4 \times 10^{-8} \mathrm{~T}$$

$$12 \times 10^{-8} \mathrm{~T}$$

JEE Main 2024 (Online) 5th April Evening Shift

MCQ (Single Correct Answer)

-1

The electrostatic force $$\left(\vec{F_1}\right)$$ and magnetic force $$\left(\vec{F}_2\right)$$ acting on a charge $$q$$ moving with velocity $$v$$ can be written :

$$\vec{F}_1=q \vec{B}, \vec{F}_2=q(\vec{B} \times \vec{v})$$

$$\vec{F}_1=q \vec{V} \cdot \vec{E}, \vec{F}_2=q(\vec{B} \cdot \vec{V})$$

$$\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B})$$

$$\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{B} \times \vec{V})$$

JEE Main 2024 (Online) 5th April Morning Shift

MCQ (Single Correct Answer)

-1

In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :

inside the outer conductor

outside the cable

in between the two conductors

inside the inner conductor