Magnetic Field due to a Current Element

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

The magnetic induction at the centre O in the figure shown is:

49 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 $μ T$ . the equal current flowing in the two conductors is:

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_{P} = \frac{μ_{0} I}{2 π r} \times 2$

$\Rightarrow 300 \times 1 0^{- 6} = \frac{4 π \times 1 0^{- 7}}{2 π \times 4 \times 1 0^{- 2}} \times 2$ &

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is

1 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

When the ring rotates about its axis with a uniform frequency fHz, the current flowing in the ring is

I=q/T=qf

Magnetic field at the centre of the ring is

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

AB µ r² long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be:

4 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

for $π R^{2} A r e a \to' l'$ current\

1 unit Area ® $\frac{l}{π R^{2}}$

$π r^{2} \to \frac{l}{π R^{2}} \times π r^{2}$

$i = \frac{l r^{2}}{R^{2}}$

Now, consider Amperian loop of radius small ‘r’ ln Amperian loop magnetic field will be tangential to the amperian loop.

$? \vec{B} . d \vec{i} = μ_{0} l_{e n c l o s e d}$

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by:

57 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

${\overset{?}{B}}_{P} = {\overset{?}{B}}_{upper wire} \otimes + {\overset{?}{B}}_{semi-circle} ? + {\overset{?}{B}}_{lower-wire} \otimes$

$B_{P} = - \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} - \frac{μ_{0} i}{4 π R} = \frac{μ_{0} i}{4 R} (1 - \frac{2}{π})$ pointing away from the page

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.

8 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

In external magnetic field, a magnetic force acts on every small part of the loop in direction perpendicular to the wire. Thus, loop assumes a shape (circular) in which it covers maximum area

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle?

39 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

The magnetic field B is calculated as:
B = 2 * (μ? I / 4πr) + (μ? I / 4πr) = (μ? I / 4πr) * (2 + π)

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is:

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

BA = μ?Iθ / 4πR
⇒ BA/BB = IAθARB / IBθBRA
⇒ (2(3π/2)(4)) / (3(π/3)(2))
⇒ 6/5

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

A wire $A$ , bent in the shape of an arc of a circle, carrying a current of $2 A$ and having radius $2 c m$ and another wire $B$ , also bent in the shape of arc of a circle, carrying a current of $3 A$ and having radius of $4 c m$ , are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and B at the common centre 0 is:

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$

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New answer posted•a year ago

Magnetic Field due to a Current Element Class 12th

Figure A and B show long straight wires of circular cross section (a and b with a < b), carrying current I which is uniformly distributed across the cross section. The magnitude of magnetic field B varies with radius r and can be represented as:

33 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

As b > a
The magnetic field inside the wire (rR) is B = µ? I/ (2πr).
For wire with radius a, B increases linearly to r=a, then decreases. For wire with radius b, B increases linearly to r=b, then decreases. Since a⇒ B? > B?
B? = µ? I / 2πa
B? = µ? I / 2πb
(Note: The question is likely asking for the gr

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Magnetic Field due to a Current Element

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