The figure shows two coaxial circular loops 1 and 2-which forms same solid angle - at point O-If B1 and B2 are the magnetic fields produced at the point O due to loop 1 and 2 respectively- then -

The correct option is A B_1/B_2=2 B_1=μ_oIR^22(R^2+x^2)^3/2 B_2=μ_oI(2R)^22((2R)^2+(2x)^2)^3/2 B_1B_2=2 ...

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Byju's Answer

Standard XII

Physics

Magnetic Field Due to a Circular Ring on the Axis

The figure sh...

Question

The figure shows two coaxial circular loops 1 and 2,which forms same solid angle $θ$ at point O.If $B_{1}$ and $B_{2}$ are the magnetic fields produced at the point O due to loop 1 and 2 respectively, then :

\frac{B_{1}}{B_{2}} = 1

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\frac{B_{1}}{B_{2}} = 2

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\frac{B_{1}}{B_{2}} = 8

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\frac{B_{1}}{B_{2}} = 4

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Solution

The correct option is A $\frac{B_{1}}{B_{2}} = 2$
$B_{1} = \frac{μ_{o} I R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$
$B_{2} = \frac{μ_{o} I {(2 R)}^{2}}{2 {({(2 R)}^{2} + {(2 x)}^{2})}^{3 / 2}}$
$\frac{B_{1}}{B_{2}} = 2$

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Similar questions

Q. The figure shows two coaxial circular loops

1

and

2

carrying equal current

I

, form the same solid angle

θ

at point

O

. If

B_{1}

and

B_{2}

are the magnetic fields produced at point

O

due to loop

1

and

2

respectively, then,

Q. In the loops shown in the figure, all curved section are either semicircles or quarter circles. All the loops carry the same current. The magnetic fields at the centres have magnitudes

B_{1}, B_{2}, B_{3}

and

B_{4}

respectively. Then,

Q. If

B_{1}, B_{2}

and

B_{3}

are the magnetic fields at points

2 R, R

and

\frac{R}{2}

from the axis respectively, then

B_{1}, B_{2}

and

B_{3}

are in the respective ratio of

Q. If the points

(a_{1}, b_{1})

(a_{2}, b_{2})

and

(a_{1} + a_{2}, b_{1} + b_{2})

are collinear, then show that

a_{1} b_{2} = a_{2} b_{1}

Q. Let

b_{1}, b_{2}, . . . . .

be a geometric sequence such that

b_{1} + b_{2} = 1

and

\infty \sum k = 1 b_{k} = 2

.
Given that

b_{2} < 0

, then the value of

b_{1}

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The figure shows two coaxial circular loops 1 and 2,which forms same solid angle θ at point O.If B1 and B2 are the magnetic fields produced at the point O due to loop 1 and 2 respectively, then :

The correct option is A B1B2=2B1=μoIR22(R2+x2)3/2B2=μoI(2R)22((2R)2+(2x)2)3/2B1B2=2

The figure shows two coaxial circular loops 1 and 2,which forms same solid angle $θ$ at point O.If $B_{1}$ and $B_{2}$ are the magnetic fields produced at the point O due to loop 1 and 2 respectively, then :

The correct option is A $\frac{B_{1}}{B_{2}} = 2$
$B_{1} = \frac{μ_{o} I R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$
$B_{2} = \frac{μ_{o} I {(2 R)}^{2}}{2 {({(2 R)}^{2} + {(2 x)}^{2})}^{3 / 2}}$
$\frac{B_{1}}{B_{2}} = 2$