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 2 isA- 0-2B- 6C- 1D- 5

The correct option is C 1B=μ_0/4 π.θ i/r ⇒ B ∝θ i ⇒B_1/B_2=θ_1/θ_2×i_1/i_2 Also i_1/i_2=l_2/l_1=θ_2/θ_1 Hence B_1/B_2=1/1 ...

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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 AOC = $60^{\circ}$ . If $B_{1} a n d B_{2}$ 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
$B = \frac{μ_{0}}{4 π} . \frac{θ i}{r}$
$\Rightarrow B \propto θ i$
$\Rightarrow \frac{B_{1}}{B_{2}} = \frac{θ_{1}}{θ_{2}} \times \frac{i_{1}}{i_{2}}$
$A l s o \frac{i_{1}}{i_{2}} = \frac{l_{2}}{l_{1}} = \frac{θ_{2}}{θ_{1}} H e n c e \frac{B_{1}}{B_{2}} = \frac{1}{1}$

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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 1B=μ04π.θir ⇒B∝θi ⇒B1B2=θ1θ2×i1i2 Also i1i2=l2l1=θ2θ1 Hence B1B2=11

A cell is connected between the points A and C of a circular conductor ABCD of centre 'O' with angle AOC = $60^{\circ}$ . If $B_{1} a n d B_{2}$ 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