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Fleming's Left Hand Rule

23.If B1; B2;...

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23.If B1; B2; and B3 are the magnetic field due to I1; I2; and I3; then in ampere's circuital law B is

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_{1},_{2}

_{3}

are the magnetic field due to the wires carrying current

I_{1}, I_{2}

I_{3}

, then in the expression of Ampere's circuital law,

\oint \to B . \to d l = μ_{0} I

\to B

B_{1}, B_{2}

B_{3}

are the magnetic fields at points

2 R, R

\frac{R}{2}

from the axis respectively, then

B_{1}, B_{2}

B_{3}

are in the respective ratio of

Q. Two long parallel wires A and B separated by a distance d, carry currents

i_{1}

i_{2}

respectively in the same direction. Write the following steps in a sequential order to find the magnitude of the resultant magnetic field at a point 'P', which is between the wires and is a distance '

x

' from the wire A.
(All the physical quantities are measured in SI units)
(a) Note the given values of

i_{1}, i_{2}

d

x

.
(b) Write the formula to find the magnetic field due to a long straight current carrying wire i.e.

B = \frac{μ_{0} i}{2 π r}

(c) Find the directions of the magnetic field at 'P' due to two wires A and B, using right hand thumb rule.
(d) Determine the magnetic field at P due to wire A, using

B_{1} = \frac{μ_{0} i_{1}}{2 π x}

(e) If the directions of magnetic field are same, then the resultant magnitude is equal to the sum of

B_{1}

B_{2}

.
(f) Determine the magnetic field

B_{2}

due to wire B at point P, ie.

B_{2} = \frac{μ_{o} i_{2}}{2 π (d - x)}

(g) If the directions of magnetic fields are opposite to each other, then the resultant magnitude is equal to the difference of

B_{1}

B_{2}

.

Q. A coil of inductance

5 mH

and negligible resistance is connected to an oscillator giving an output voltage

E = 10 sin ω t

. The peak currents in the circuit for

ω = 10 s^{- 1},

100 s^{- 1}

500 s^{- 1}

I_{1}, I_{2}

I_{3}

respectively. Then the ratio

I_{1} : I_{2} : I_{3}

Q. Three current carrying elements having same length carries current

i, 2 i

4 i

respectively. Magnetic fields due to wires be

B_{1}, B_{2}

B_{3}

respectively at a point

P

whose distance from the current elements is

r, 2 r

4 r

respectively. Then,

B_{1} : B_{2} : B_{3}

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