- 1(current)
Q. No.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 \(10~\text{cm}.\) If the magnetic field is uniform and has a value \(10^{-2}~ \text{T},\) then the flux through the loop will be:
| 1. | \(2 \pi \times 10^{-2}~\text{Wb}\) | 2. | \(3 \pi \times 10^{-4}~\text{Wb}\) |
| 3. | \(5 \pi \times 10^{-5}~\text{Wb}\) | 4. | \(5 \pi \times 10^{-4}~\text{Wb}\) |
What is the dimensional formula of magnetic flux?
1. \(\left[ M L^2 T^{-2}A^{-1}\right]\)
2. \(\left[ M L^1 T^{-1}A^{-2}\right]\)
3. \(\left[ M L^2 T^{-3}A^{-1}\right]\)
4. \(\left[ M L^{-2} T^{-2}A^{-2}\right]\)
- 1(current)
Q. No.
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