Application of Ampere's Law : Magnetic Field Due to Circular Conductor
Trending Questions
Q. A circular conductor of radius ro=1cm has an internal field,
¯¯¯¯¯H=104ro(1a2sinaro−roacosaro)ˆaϕ(A/m)
where, a=π2r0
¯¯¯¯¯H=104ro(1a2sinaro−roacosaro)ˆaϕ(A/m)
where, a=π2r0
- 8π Ampere
- 8πAmpere
- 4π Ampere
- 4πAmpere
Q. An infinitely long uniform solid wire of radius a carries a uniform dc current of density →j.
A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.
The magnetic field inside the hole is
A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.
The magnetic field inside the hole is
- uniform and depends only on d
- uniform and depends only on b
- uniform and depends on both b and d
- non uniform
Q. The magnetic field intensity →H at the center of the current of the carrying circular loops shown in the figure below is
- −IR^az
- −I2R^az
- IR^az
- I2R^az
Q. The current diensty in a medium is given by
→J=400sinθ2π(r2+4)^arAm−2
The total current and the average current density flowing through the portion of a spherical surface
r = 0.8 m, π12≤θ≤π4, 0≤2π are given, respectively, by
→J=400sinθ2π(r2+4)^arAm−2
The total current and the average current density flowing through the portion of a spherical surface
r = 0.8 m, π12≤θ≤π4, 0≤2π are given, respectively, by
- 15.09 A, 12.86 Am−2
- 7.17 A, 6.88 Am−2
- 12.86 A, 9.23 Am−2
- 10.28 A, 7.56 Am−2
Q. The magnetic field intensity →H at the centre of current carrying loop (circular in space) and kept in x -y plane as shown will be
- I2R^az
- −I2R^az
- IR^az
- −IR^az