A tightly wound \(100\) turns coil of radius \(10~\text{cm}\) carries a current of \(7~\text A\). The magnitude of the magnetic field at the centre of the coil is: (Take permeability of free space as \(​4 \pi \times 10^{-7​}\)SI units):
1. \(4.4~\text T\)
2. \(4.4~\text {mT}\)
3. \(44~\text T\)
4. \(44~\text {mT}\)
Subtopic:  Magnetic Field due to various cases |
 71%
Level 2: 60%+
NEET - 2024
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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 |
 57%
Level 3: 35%-60%
NEET - 2023
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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
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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 = \(4 \pi \times 10^{-7}~ \text{H/m}\))
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
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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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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 long solenoid of \(50~\text{cm}\) length having \(100\) turns carries a current of \(2.5~\text{A}\). The magnetic field at the centre of the solenoid is: 
\(\big(\mu_0 = 4\pi\times 10^{-7}~\text{TmA}^{-1} \big)\)
1. \(3.4\times 10^{-4}~\text{T}\)
2. \(6.28\times 10^{-5}~\text{T}\)
3. \(3.14\times 10^{-5}~\text{T}\)
4. \(6.28\times 10^{-4}~\text{T}\)

Subtopic:  Magnetic Field due to various cases |
 69%
Level 2: 60%+
NEET - 2020
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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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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. The magnetic field at the centre of this coil of \(n\) turns will be:
1. \(nB\)
2. \(n^2B\)
3. \(2nB\)
4. \(2n^2B\)
Subtopic:  Magnetic Field due to various cases |
 85%
Level 1: 80%+
NEET - 2016
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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 |
 67%
Level 2: 60%+
NEET - 2015
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