A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle AOC -60- If B 1 and B 2 are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively- the ratio B 1- B 3 isA- 0-2B- 6C- 5D- 1

The correct option is D 1 B=μ _0/4πθ i/4⇒ B∝θ i ( but i_1/i_2=l_2/l_1=θ _2/θ _1 ) ⇒ B_1/B_2=θ _1/θ _2.i_1/i_2 So, B_1/B_2=θ _1/θ _2×θ _2/θ _1 ⇒ B_1=B_2 ...

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Byju's Answer

Standard XII

Physics

Magnetic Field

A cell is con...

Question

A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle $A O C = 60^{\circ}$ . If $B_{1}$ and $B_{2}$ are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio $\frac{B_{1}}{B_{2}}$ is

0.2

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Solution

The correct option is C 1

$B = \frac{μ_{0}}{4 π} \frac{θ i}{4} \Rightarrow B \propto θ i$ $(b u t \frac{i_{1}}{i_{2}} = \frac{l_{2}}{l_{1}} = \frac{θ_{2}}{θ_{1}})$
$\Rightarrow \frac{B_{1}}{B_{2}} = \frac{θ_{1}}{θ_{2}} . \frac{i_{1}}{i_{2}}$
So, $\frac{B_{1}}{B_{2}} = \frac{θ_{1}}{θ_{2}} \times \frac{θ_{2}}{θ_{1}}$
$\Rightarrow B_{1} = B_{2}$

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A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle AOC=60∘. If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio B1B2 is

The correct option is C 1 B=μ04πθi4⇒B∝θi (but i1i2=l2l1=θ2θ1) ⇒ B1B2=θ1θ2.i1i2 So, B1B2=θ1θ2×θ2θ1 ⇒ B1=B2

A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle $A O C = 60^{\circ}$ . If $B_{1}$ and $B_{2}$ are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio $\frac{B_{1}}{B_{2}}$ is