The net magnetic flux through any closed surface, kept in uniform magnetic field is:
1. zero
2. \(\frac{\mu_{0}}{4 \pi}\)
3. \(4\pi μ_{0}\)
4. \(\frac{4\mu_{0}}{\pi}\)
A circular disc of radius \(0.2\) m is placed in a uniform magnetic field of induction \(\frac{1}{\pi} \left(\frac{\text{Wb}}{\text{m}^{2}}\right)\)
in such a way that its axis makes an angle of \(60^{\circ}\) with \(\vec {B}.\) The magnetic flux linked to the disc will be:
1. \(0.02\) Wb
2. \(0.06\) Wb
3. \(0.08\) Wb
4. \(0.01\) Wb
If a current is passed through a circular loop of radius \(R\) then magnetic flux through a coplanar square loop of side \(l\) as shown in the figure \((l<<R)\) is:
1. \(\frac{\mu_{0} l}{2} \frac{R^{2}}{l}\)
2. \(\frac{\mu_{0} I l^{2}}{2 R}\)
3. \(\frac{\mu_{0} l \pi R^{2}}{2 l}\)
4. \(\frac{\mu_{0} \pi R^{2} I}{l}\)
The radius of a loop as shown in the figure is \(10~\mathrm {cm}.\) If the magnetic field is uniform and has a value \(10^{-2}~ T,\) then the flux through the loop will be:
1. | \(2 \pi \times 10^{-2}Wb\) | 2. | \(3 \pi \times 10^{-4}Wb\) |
3. | \(5 \pi \times 10^{-5}Wb\) | 4. | \(5 \pi \times 10^{-4}Wb\) |
A square of side L meters lies in the XY-plane in a region where the magnetic field is given by \(\vec{B}=B_{0}\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )~T\) where \(B_{0}\) is constant. The magnitude of flux passing through the square will be:
1.
2.
3.
4.
What is the dimensional formula of magnetic flux?
1.
2.
3.
4.
1. | \(0\) | 2. | \(2\) weber |
3. | \(0.5\) weber | 4. | \(1\) weber |
A circular loop of radius R carrying current i lies in the x-y plane. If the centre of the loop coincides with the origin, then the total magnetic flux passing through the x-y plane will be:
1. | directly proportional to I. |
2. | directly proportional to R. |
3. | directly proportional to R2. |
4. | Zero. |