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Q. No.Which one of the following gives the value of the magnetic field according to Biot-Savart’s law?
| 1. | \(\dfrac{{i} \Delta {l} \sin (\theta)}{{r}^2} \) | 2. | \(\dfrac{\mu_0}{4 \pi} \dfrac{i \Delta {l} \sin (\theta)}{r} \) |
| 3. | \(\dfrac{\mu_0}{4 \pi} \dfrac{{i} \Delta{l} \sin (\theta)}{{r}^2} \) | 4. | \(\dfrac{\mu_0}{4 \pi} {i} \Delta {l} \sin (\theta)\) |
Subtopic: Biot-Savart Law |
90%
From NCERT
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To maximise the magnetic field caused by a small element of a current-carrying conductor at a point, the angle between the element and the line connecting the element to the point \(P\) must be:
| 1. | \(0^{\circ}\) | 2. | \(90^{\circ}\) |
| 3. | \(180^{\circ}\) | 4. | \(45^{\circ}\) |
Subtopic: Biot-Savart Law |
89%
From NCERT
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An element \(\Delta l=\Delta x \hat{i}\) is placed at the origin and carries a large current of \(I=10~\text A\) (as shown in the figure). What is the magnetic field on the \(y\text-\)axis at a distance of \(0.5~\text m?\)
\((\text{Given}~\Delta x=1~\text{cm})\)
| 1. | \(6\times 10^{-8}~\text{T}\) | 2. | \(4\times 10^{-8}~\text{T}\) |
| 3. | \(5\times 10^{-8}~\text{T}\) | 4. | \(5.4\times 10^{-8}~\text{T}\) |
Subtopic: Biot-Savart Law |
81%
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A straight wire carrying a current of \(12~\text{A}\) is bent into a semi-circular arc of radius \(2.0~\text{cm}\) as shown in the figure. Considering the magnetic field \(B\) at the centre of the arc, what will be the magnetic field due to the straight segments?
| 1. | \(0\) | 2. | \(1.2\times 10^{-4}~\text{T}\) |
| 3. | \(2.1\times 10^{-4}~\text{T}\) | 4. | None of these |
Subtopic: Biot-Savart Law |
80%
From NCERT
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Which one of the following expressions represents Biot-Savart's law? Symbols have their usual meanings.
| 1. | \(\overrightarrow{d B}=\frac{\mu_0 \mathrm{I}(\overrightarrow{d l} \times \hat r)}{4 \pi|\overrightarrow{\mathrm{r}}|^3}\\ \) | 2. | \(\overrightarrow{d B}=\frac{\mu_0 \mathrm{I}(\overrightarrow{d l} \times \hat r)}{4 \pi|\overrightarrow{\mathrm{r}}|^2} \) |
| 3. | \(\overrightarrow{d B}=\frac{\mu_0 \mathrm{I}(\overrightarrow{d l} \times \vec{r})}{4 \pi|\vec{r}|^3} \) | 4. | \(\overrightarrow{d B}=\frac{\mu_0 \mathrm{I}(\overrightarrow{d l} \cdot \vec{r})}{4 \pi|\overrightarrow{\mathrm{r}}|^3}\) |
Subtopic: Biot-Savart Law |
69%
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A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is \(B\). It is then bent into a circular coil of \(n\) turns. What will the magnetic field be at the centre of this \(n\text-\)turn coil?
| 1. | \(nB\) | 2. | \(n^2B\) |
| 3. | \(2nB\) | 4. | \(2n^2B\) |
Subtopic: Magnetic Field due to various cases |
84%
From NCERT
NEET - 2016
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The magnetic induction at point \(P\), which is \(4\) cm from a long current-carrying wire is \(10^{-8}\) Tesla. What would be the field of induction at a distance of \(12\) cm from the same current?
| 1. | \(3.33\times 10^{-9}\) Tesla |
| 2. | \(1.11\times 10^{-4}\) Tesla |
| 3. | \(3\times 10^{-3}\) Tesla |
| 4. | \(9\times 10^{-2}\) Tesla |
Subtopic: Magnetic Field due to various cases |
81%
From NCERT
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Two similar coils of radius \(R\) are lying concentrically with their planes at right angles to each other. The currents flowing in them are \(I\) and \(2I,\) respectively. What will be the resultant magnetic field induction at the centre?
| 1. | \(\sqrt{5} \mu_0I \over 2R\) | 2. | \({3} \mu_0I \over 2R\) |
| 3. | \( \mu_0I \over 2R\) | 4. | \( \mu_0I \over R\) |
Subtopic: Magnetic Field due to various cases |
82%
From NCERT
AIPMT - 2012
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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. | \(0\) |
Subtopic: Magnetic Field due to various cases |
82%
From NCERT
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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. |  |
Subtopic: Magnetic Field due to various cases |
80%
From NCERT
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