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Q. No.Given below are two statements:
| Statement I: | Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element \((Idl)\) of a current-carrying conductor only. |
| Statement II: | Biot-Savart's law is analogous to Coulomb's inverse square law of charge \(q,\) with the former being related to the field produced by a scalar source, \((Idl)\) while the latter being produced by a vector source, \(q.\) |
| 1. | Statement I is incorrect but Statement II is correct. |
| 2. | Both Statement I and Statement II are correct. |
| 3. | Both Statement I and Statement II are incorrect. |
| 4. | Statement I is correct but Statement II is incorrect. |
Subtopic: Biot-Savart Law |
56%
Level 3: 35%-60%
NEET - 2022
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A very long conducting wire is bent in a semi-circular shape from \(A\) to \(B\) as shown in the figure. The magnetic field at the point \(P\) for steady current configuration is given by:
| 1. | \(\dfrac{\mu_0 i}{4 R}\left[1-\dfrac{2}{\pi}\right]\) pointed into the page |
| 2. | \(\dfrac{\mu_0 i}{4 R}\) pointed into the page |
| 3. | \(\dfrac{\mu_0 i}{4 R}\) pointed away from the page |
| 4. | \(\dfrac{\mu_0 i}{4 R}\left[1-\dfrac{2}{\pi}\right]\) pointed away from the page |
Subtopic: Magnetic Field due to various cases |
58%
Level 3: 35%-60%
NEET - 2023
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The magnetic field on the axis of a circular loop of radius \(100~\text{cm}\) carrying current \(I=\sqrt{2}~\text{A},\) at a point \(1~\text{m}\) away from the centre of the loop is given by:
| 1. | \(3.14 \times 10^{-7} ~\text{T} \) | 2. | \(6.28 \times 10^{-7} ~\text{T} \) |
| 3. | \(3.14 \times 10^{-4} ~\text{T} \) | 4. | \(6.28 \times 10^{-4} ~\text{T}\) |
Subtopic: Magnetic Field due to various cases |
58%
Level 3: 35%-60%
NEET - 2022
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A closely packed coil having \(1000\) turns has an average radius of \(62.8 ~\text{cm}\). If the current carried by the wire of the coil is \(1~ \text{A},\) the value of the magnetic field produced at the centre of the coil will be nearly:
(permeability of free space = \(4 \pi \times 10^{-7}~ \text{H/m}\))
(permeability of free space = \(4 \pi \times 10^{-7}~ \text{H/m}\))
| 1. | \(10^{-1}~\text{T}\) | 2. | \(10^{-2}~\text T\) |
| 3. | \(10^{2}~\text T\) | 4. | \(10^{-3}~\text{T}\) |
Subtopic: Magnetic Field due to various cases |
67%
Level 2: 60%+
NEET - 2022
Hints
The shape of the magnetic field lines due to an infinite long, straight current carrying conductor is:
| 1. | a straight line | 2. | circular |
| 3. | elliptical | 4. | a plane |
Subtopic: Magnetic Field due to various cases |
72%
Level 2: 60%+
NEET - 2022
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A long solenoid of radius \(1~\text{mm}\) has \(100\) turns per mm. If \(1~\text{A}\) current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
| 1. | \(6.28 \times 10^{-4} ~\text{T} \) | 2. | \(6.28 \times 10^{-2}~\text{T}\) |
| 3. | \(12.56 \times 10^{-2}~\text{T}\) | 4. | \(12.56 \times 10^{-4} ~\text{T}\) |
Subtopic: Magnetic Field due to various cases |
65%
Level 2: 60%+
NEET - 2022
Hints
A straight conductor carrying current \(I\) splits into two parts as shown in the figure. The radius of the circular loop is \(R\). The total magnetic field at the centre \(P\) of the loop is:
| 1. | zero | 2. | \(\dfrac{3\mu_0 i}{32R},~\text{inward}\) |
| 3. | \(\dfrac{3\mu_0 i}{32R},~\text{outward}\) | 4. | \(\dfrac{\mu_0 i}{2R},~\text{inward}\) |
Subtopic: Magnetic Field due to various cases |
78%
Level 2: 60%+
NEET - 2019
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An electron moving in a circular orbit of radius \(r\) makes \(n\) rotations per second. The magnetic field produced at the centre has a magnitude:
| 1. | \(\dfrac{\mu_0ne}{2\pi r}\) | 2. | zero |
| 3. | \(\dfrac{n^2e}{r}\) | 4. | \(\dfrac{\mu_0ne}{2r}\) |
Subtopic: Magnetic Field due to various cases |
69%
Level 2: 60%+
NEET - 2015
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A wire carrying current \(I\) has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to \(X\)-axis while the semicircular portion of radius \(R\) is lying in the \(Y\text-Z\) plane. The magnetic field at point \(O\) is:
1. \(B=\frac{\mu i }{4\pi R}\left ( \pi \hat{i}+2\hat{k} \right )\)
2. \(B=-\frac{\mu i }{4\pi R}\left ( \pi \hat{i}-2\hat{k} \right )\)
3. \(B=-\frac{\mu i }{4\pi R}\left ( \pi \hat{i}+2\hat{k} \right )\)
4. \(B=\frac{\mu i }{4\pi R}\left ( \pi \hat{i}-2\hat{k} \right )\)
Subtopic: Magnetic Field due to various cases |
68%
Level 2: 60%+
NEET - 2015
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Two identical long conducting wires \(({AOB})\) and \(({COD})\) 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. The 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. | \(\dfrac{\mu_0}{2\pi d}\sqrt{\left[I^2_1+I^2_2\right ]}\) |
Subtopic: Magnetic Field due to various cases |
77%
Level 2: 60%+
AIPMT - 2014
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