Q. No.Which one of the following gives the value of the magnetic field according to Biot-Savart’s law?
| 1. | \(\frac{{i} \Delta {l} \sin (\theta)}{{r}^2} \) | 2. | \(\frac{\mu_0}{4 \pi} \frac{i \Delta {l} \sin (\theta)}{r} \) |
| 3. | \(\frac{\mu_0}{4 \pi} \frac{{i} \Delta{l} \sin (\theta)}{{r}^2} \) | 4. | \(\frac{\mu_0}{4 \pi} {i} \Delta {l} \sin (\theta)\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
An element \(\Delta l=\Delta x \hat{i}\) is placed at the origin and carries a large current of \(I=10\) A (as shown in the figure). What is the magnetic field on the \(y\text-\)axis at a distance of \(0.5\) m? \((\Delta x=1~\text{cm})\)
| 1. | \(6\times 10^{-8}~\text{T}\) | 2. | \(4\times 10^{-8}~\text{T}\) |
| 3. | \(5\times 10^{-8}~\text{T}\) | 4. | \(5.4\times 10^{-8}~\text{T}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
| 1. | \(0\) | 2. | \(1.2\times 10^{-4}~\text{T}\) |
| 3. | \(2.1\times 10^{-4}~\text{T}\) | 4. | None of these |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
The resistances of three parts of a circular loop are as shown in the figure. What will be the magnetic field at the centre of \(O\)
(current enters at \(A\) and leaves at \(B\) and \(C\) as shown)?
| 1. | \(\dfrac{\mu_{0} I}{6 a}\) | 2. | \(\dfrac{\mu_{0} I}{3 a}\) |
| 3. | \(\dfrac{2\mu_{0} I}{3 a}\) | 4. | \(0\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Which of the following graphs correctly represents the variation of magnetic field induction with distance due to a thin wire carrying current?
| 1. |  |
2. |  |
| 3. |  |
4. |  |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
What is the magnetic field at point \(O\) in the figure?
| 1. | \(\dfrac{\mu_{0} I}{4 \pi r}\) | 2. | \(\dfrac{\mu_{0} I}{4 \pi r} + \dfrac{\mu_{0} I}{2 \pi r}\) |
| 3. | \(\dfrac{\mu_{0} I}{4 r} + \dfrac{\mu_{0} I}{4 \pi r}\) | 4. | \(\dfrac{\mu_{0} I}{4 r} - \dfrac{\mu_{0} I}{4 \pi r}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Two identical long conducting wires \(\mathrm{AOB}\) and \(\mathrm{COD}\) are placed at a right angle to each other, with one above the other such that '\(O\)' is the common point for the two. The wires carry \(I_1\) and \(I_2\) currents, respectively. Point '\(P\)' is lying at a distance '\(d\)' from '\(O\)' along a direction perpendicular to the plane containing the wires. What will be the magnetic field at the point \(P\)?
| 1. | \(\dfrac{\mu_0}{2\pi d}\left(\dfrac{I_1}{I_2}\right )\) | 2. | \(\dfrac{\mu_0}{2\pi d}\left[I_1+I_2\right ]\) |
| 3. | \(\dfrac{\mu_0}{2\pi d}\left[I^2_1+I^2_2\right ]\) | 4. | \(\dfrac{\mu_0}{2\pi d}\sqrt{\left[I^2_1+I^2_2\right ]}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
(\(R\) = radius of the coil)
| 1. | \(R \over 3\) | 2. | \(\sqrt{3}R\) |
| 3. | \(R \over \sqrt3\) | 4. | \(R \over 2\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
| 1. |  |
2. |  |
| 3. |  |
4. |  |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
| 1. | \( \dfrac{\mu_0 i}{2 \sqrt{2} R} \) | 2. | \( \dfrac{\mu_0 i}{2 R} \) |
| 3. | \( \dfrac{\mu_0 i}{4 R} \) | 4. | \( \dfrac{\mu_0 i}{\sqrt{2} R}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
© 2024 GoodEd Technologies Pvt. Ltd.