As shown in the figure, a current of 2 A flowing in an equilateral triangle of side cm. The magnetic field at the centroid of the triangle is [2023]

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Q.
As shown in the figure, a current of 2 A flowing in an equilateral triangle of side $4 \sqrt{3}$ cm. The magnetic field at the centroid $O$ of the triangle is [2023]

(Neglect the effect of earth's magnetic field.)

1 $4 \sqrt{3} \times 10^{- 4} T$

2 $4 \sqrt{3} \times 10^{- 5} T$

3 $\sqrt{3} \times 10^{- 4} T$

4 $3 \sqrt{3} \times 10^{- 5} T$

Ans.
(4)

$d \tan 60 ° = 2 \sqrt{3}$

$d = 2 cm$

$B = 3 \times \frac{μ_{0} i}{2 π d} \sin 60 °$

$= 3 \times \frac{2 \times 10^{- 7} \times 2}{2 \times 10^{- 2}} \times \frac{\sqrt{3}}{2}$

$= 3 \sqrt{3} \times 10^{- 5}$

Solution

Q.
As shown in the figure, a current of 2 A flowing in an equilateral triangle of side $4 \sqrt{3}$ cm. The magnetic field at the centroid $O$ of the triangle is [2023]

(Neglect the effect of earth's magnetic field.)

1 $4 \sqrt{3} \times 10^{- 4} T$

2 $4 \sqrt{3} \times 10^{- 5} T$

3 $\sqrt{3} \times 10^{- 4} T$

4 $3 \sqrt{3} \times 10^{- 5} T$

Ans.
(4)

$d \tan 60 ° = 2 \sqrt{3}$

$d = 2 cm$

$B = 3 \times \frac{μ_{0} i}{2 π d} \sin 60 °$

$= 3 \times \frac{2 \times 10^{- 7} \times 2}{2 \times 10^{- 2}} \times \frac{\sqrt{3}}{2}$

$= 3 \sqrt{3} \times 10^{- 5}$

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Solution

Q. As shown in the figure, a current of 2 A flowing in an equilateral triangle of side 43 cm. The magnetic field at the centroid O of the triangle is [2023] (Neglect the effect of earth's magnetic field.)

1 43×10-4 T

2 43×10-5 T

3 3×10-4 T

4 33×10-5 T

Ans. (4) d tan 60°=23 d=2 cm B=3×μ0i2πdsin60° =3×2×10-7×22×10-2×32 =33×10-5

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Q.
As shown in the figure, a current of 2 A flowing in an equilateral triangle of side $4 \sqrt{3}$ cm. The magnetic field at the centroid $O$ of the triangle is [2023]

(Neglect the effect of earth's magnetic field.)

1 $4 \sqrt{3} \times 10^{- 4} T$

2 $4 \sqrt{3} \times 10^{- 5} T$

3 $\sqrt{3} \times 10^{- 4} T$

4 $3 \sqrt{3} \times 10^{- 5} T$

Ans.
(4)

$d \tan 60 ° = 2 \sqrt{3}$

$d = 2 cm$

$B = 3 \times \frac{μ_{0} i}{2 π d} \sin 60 °$

$= 3 \times \frac{2 \times 10^{- 7} \times 2}{2 \times 10^{- 2}} \times \frac{\sqrt{3}}{2}$

$= 3 \sqrt{3} \times 10^{- 5}$