English

A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field

Advertisements
Advertisements

Question

A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is ______.

Options

  • `1/2`

  • 1

  • 2

  • 4

MCQ
Fill in the Blanks
Advertisements

Solution

A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is 1.

Explanation:

Consider two amperian loops of radius `a/2` and 2a. Applying Ampere's circuital law for these loops.

We get `ointB.dL = μ_0I_{"enclosed"}`

For the smaller loop,

`B_1 xx 2pi a/2 = mu_0 xx pia^2 1 xx (a/2)^2`

`mu_0I xx 1/4 = (mu_0I)/4`

`B_1 = (mu_0I)/(4pia)`

`B_2 xx 2pi(2a) = mu_0I`

`B_2 = (mu_0I)/(4pia)`

∴ B1/B2 = `(mu_0I)/(4pia) xx (4pia)/(mu_0I) = 1`

shaalaa.com
  Is there an error in this question or solution?
2022-2023 (March) Outside Delhi Set 1

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?


A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across the cross-section. Apply Ampere’s circuital law to calculate the magnetic field at a point ‘r’ in the region for (i) r < a and (ii) r > a.


In Ampere's  \[\oint \vec{B}  \cdot d \vec{l}  =  \mu_0 i,\] the current outside the curve is not included on the right hand side. Does it mean  that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve?  


A long, straight wire carries a current. Is Ampere's law valid for a loop that does not enclose the wire, or that encloses the wire but is not circular?


In order to have a current in a long wire, it should be connected to a battery or some such device. Can we obtain the magnetic due to a straight, long wire by using Ampere's law without mentioning this other part of the circuit? 


A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross section. Find the magnetic field B at a point at a distance (a) 2 cm (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph B versus x for 0 < x < 20 cm. 


Find the magnetic field due to a long straight conductor using Ampere’s circuital law.


The force required to double the length of a steel wire of area 1 cm2, if it's Young's modulus Y = `2 xx 10^11/m^2` is:


A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns If it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid is:


Read the following paragraph and answer the questions.

Consider the experimental set-up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting non-magnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off.

  1. Explain the reason for the jumping of the ring when the switch is closed in the circuit.
  2. What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
  3. Explain the two laws that help us understand this phenomenon.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×