What is the torque on circular loop with given parameters?

5024×10⁻⁷ĵ Wb·A. m=0.04π(î+k̂). (î+k̂)×(3î+2k̂)=ĵ. τ=0.04π×4×10⁻³ĵ.

Circular loop radius 2cm n=(k̂+î)/√2 in B=B₀(3î+2k̂) B₀=4×10⁻³T I=100√2A — torque

Q: What is (î+k̂)×(3î+2k̂)?

ĵ. Expanding: 3(î×î)+2(î×k̂)+3(k̂×î)+2(k̂×k̂)=0−2ĵ+3ĵ+0=ĵ.

JEERANKERS

PhysicsMagnetism

QMCQMagnetism

A current carrying circular loop of radius 2 cm with unit normal $\hat{n}=(\hat{k}+\hat{i})/\sqrt{2}$ is placed in $\vec{B}=B_0(3\hat{i}+2\hat{k})$. $B_0=4\times10^{-3}$ T, $I=100\sqrt{2}$ A. Torque is:

A) $16\times10^{-5}\hat{k}$ B) $5024\times10^{-7}\hat{k}$ C) $5024\times10^{-7}\hat{i}$ D) $5024\times10^{-7}\hat{j}$

✅ Correct Answer

D) 5024×10⁻⁷ĵ Wb·A

✓

Solution

Compute magnetic moment m⃗

$\vec{m}=I\cdot\pi r^2\cdot\hat{n}=100\sqrt{2}\times\pi(0.02)^2\times\dfrac{\hat{i}+\hat{k}}{\sqrt{2}}=100\pi\times4\times10^{-4}(\hat{i}+\hat{k})$

$=0.04\pi(\hat{i}+\hat{k})\text{ A·m}^2$

Evaluate cross product

$(\hat{i}+\hat{k})\times(3\hat{i}+2\hat{k})=3(\hat{i}\times\hat{i})+2(\hat{i}\times\hat{k})+3(\hat{k}\times\hat{i})+2(\hat{k}\times\hat{k})$

$=0+2(-\hat{j})+3(\hat{j})+0=\hat{j}$

Compute torque

$\vec{\tau}=\vec{m}\times\vec{B}=0.04\pi(\hat{i}+\hat{k})\times4\times10^{-3}(3\hat{i}+2\hat{k})$

$=0.04\pi\times4\times10^{-3}\times\hat{j}=0.04\times3.14\times4\times10^{-3}\hat{j}$

$=5024\times10^{-7}\hat{j}$

📘 Torque on Magnetic Dipole τ = m × B

$\vec{m}=IA\hat{n}$. Note $\sqrt{2}$ in $I$ and denominator of $\hat{n}$ cancel. $\vec{\tau}=\vec{m}\times\vec{B}$. Cross product $(\hat{i}+\hat{k})\times(3\hat{i}+2\hat{k})=\hat{j}$.

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Important Concepts

Cancellation of √2I=100√2, n̂ has /√2. Product I×n̂=100√2×(î+k̂)/√2=100(î+k̂). Clean calculation!

Area A=πr²A=π×(0.02)²=4π×10⁻⁴ m². m=100×4π×10⁻⁴(î+k̂)=0.04π(î+k̂).

Cross Product Rulesî×î=0, î×k̂=−ĵ, k̂×î=ĵ, k̂×k̂=0. These are standard cyclic rules.

Numerical Resultτ=0.04π×4×10⁻³=0.04×3.14×4×10⁻³=0.5024×10⁻³=5024×10⁻⁷ Wb·A.

FAQs

How does √2 cancel?

⌄

I=100√2 and n̂=1/√2 factor. Product=100√2/√2=100.

What is (î+k̂)×(3î+2k̂)?

⌄

Expand: 0+2(î×k̂)+3(k̂×î)+0=−2ĵ+3ĵ=ĵ.

Why is torque along ĵ?

⌄

Both m and B lie in the xz-plane. Their cross product must be perpendicular to xz-plane, i.e., along ŷ=ĵ.

What is torque magnitude?

⌄

0.5024×10⁻³ N·m = 5024×10⁻⁷ Wb·A.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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