A current carrying circular loop of radius 2 cm with unit normal is placed in a magnetic field

Question

A current carrying circular loop of radius 2 cm with unit normal $\hat{n} = \frac{\hat{k} + \hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B} = B_{0} (3 \hat{i} + 2 \hat{k}) .$ If $B_{0} = 4 \times 10^{- 3} T$ and current $I = 100 \sqrt{2} A$ , the torque experienced by the loop is ______ $Wb.A$ . $(π = 3.14)$ [2026]

Smart Achievers · Accepted Answer

(4)

$\vec{τ} = I \vec{A} \times \vec{B} = I π r^{2} \hat{n} \times \vec{B}$

$\vec{τ} = 100 \sqrt{2} \times 3.14 \times {(2 \times 10^{- 2})}^{2} [\frac{\hat{k} + \hat{i}}{\sqrt{2}}] \times B_{0} (3 \hat{i} + 2 \hat{k})$

$= \frac{314 \times \sqrt{2} \times 4 \times 10^{- 4}}{\sqrt{2}} \times B_{0} (3 \hat{j} - 2 \hat{j})$

$\vec{τ} = 1256 \times 10^{- 4} \times 4 \times 10^{- 3} (\hat{j}) Wb \cdot A$

$= 5024 \times 10^{- 7} (\hat{j}) Wb \cdot A$

Solution

1 $16 \times 10^{- 5} \hat{k}$

2 $5024 \times 10^{- 7} \hat{k}$

3 $5024 \times 10^{- 7} \hat{i}$

4 $5024 \times 10^{- 7} \hat{j}$

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Solution

Q. A current carrying circular loop of radius 2 cm with unit normal n^=k^+i^2 is placed in a magnetic field, B→=B0(3i^+2k^). If B0=4×10-3 T and current I=1002 A, the torque experienced by the loop is ______ Wb.A. (π=3.14) [2026]

1 16×10-5k^

2 5024×10-7k^

3 5024×10-7i^

4 5024×10-7j^

Ans. (4) τ→=IA→×B→=Iπr2n^×B→ τ→=1002×3.14×(2×10-2)2[k^+i^2]×B0(3i^+2k^) =314×2×4×10-42×B0(3j^-2j^) τ→=1256×10-4×4×10-3(j^) Wb·A =5024×10-7(j^) Wb·A

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1 $16 \times 10^{- 5} \hat{k}$

2 $5024 \times 10^{- 7} \hat{k}$

3 $5024 \times 10^{- 7} \hat{i}$

4 $5024 \times 10^{- 7} \hat{j}$