A current carrying circular loop of radius 2 cm with unit normal n = (k + i)/sqrt(2) is placed in a magnetic field, B = Bo (3i + 2k). If Bo = 4x10^-3 T and current I = 100sqrt(2) A, the torque experienced by the loop is _______ Wb.A. (pi=3.14)

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q39 – JEE-MAIN 2026

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Home Physics Moving ChargesQuestion #349

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_{o} (3 \hat{i} + 2 \hat{k})

. If

B_{o} = 4 \times 10^{- 3}

T and current

I = 100 \sqrt{2}

A, the torque experienced by the loop is _______ Wb.A. (

π = 3.14

)

Clue 1 of 3: Understand the concept of torque on a current loop

Torque on a current loop in a magnetic field is due to the interaction between the magnetic dipole moment of the loop and the external magnetic field.

🧠 Full Solution Path

Step 1: Calculate Magnetic Dipole Moment✦ Active

First, calculate the area of the circular loop. Given radius $r = 2 cm = 2 \times 10^{- 2} m$ . The area is $A = π r^{2} = π (2 \times 10^{- 2})^{2} = 4 π \times 10^{- 4} m^{2}$ . Then, calculate the magnetic dipole moment $\vec{M} = I A \hat{n}$ . Given current $I = 100 \sqrt{2} A$ and unit normal $\hat{n} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$ . Substitute these values:

\vec{M} = (100 \sqrt{2}) (4 π \times 10^{- 4}) (\frac{\hat{i} + \hat{k}}{\sqrt{2}})

\vec{M} = (100) (4 π \times 10^{- 4}) (\hat{i} + \hat{k})

\vec{M} = 0.04 π (\hat{i} + \hat{k}) {A.m}^{2}

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before calculation.

Step 2: Calculate Torque on the Loop○ Expand

The torque experienced by the loop is given by $\vec{τ} = \vec{M} \times \vec{B}$ . Given magnetic field $\vec{B} = B_{o} (3 \hat{i} + 2 \hat{k})$ with $B_{o} = 4 \times 10^{- 3} T$ . So, $\vec{B} = (4 \times 10^{- 3}) (3 \hat{i} + 2 \hat{k}) = (12 \times 10^{- 3}) \hat{i} + (8 \times 10^{- 3}) \hat{k} T$ . Now perform the cross product:

\vec{τ} = [0.04 π (\hat{i} + \hat{k})] \times [(12 \times 10^{- 3}) \hat{i} + (8 \times 10^{- 3}) \hat{k}]

\vec{τ} = (0.04 π \times 10^{- 3}) [(\hat{i} + \hat{k}) \times (12 \hat{i} + 8 \hat{k})]

\vec{τ} = (0.04 π \times 10^{- 3}) [(\hat{i} \times 8 \hat{k}) + (\hat{k} \times 12 \hat{i})]

\vec{τ} = (0.04 π \times 10^{- 3}) [- 8 \hat{j} + 12 \hat{j}]

\vec{τ} = (0.04 π \times 10^{- 3}) (4 \hat{j})

\vec{τ} = 0.16 π \times 10^{- 3} \hat{j} N.m

💡 Teacher's Secret Hint

Remember the properties of vector cross products: $\hat{i} \times \hat{i} = 0$ , $\hat{k} \times \hat{k} = 0$ , $\hat{i} \times \hat{k} = - \hat{j}$ , and $\hat{k} \times \hat{i} = \hat{j}$ .

Step 3: Substitute Value of Pi and Final Result○ Expand

Substitute the given value of $π = 3.14$ into the torque expression:

\vec{τ} = 0.16 \times 3.14 \times 10^{- 3} \hat{j}

\vec{τ} = 0.5024 \times 10^{- 3} \hat{j}

\vec{τ} = 5024 \times 10^{- 7} \hat{j} Wb.A

Comparing this result with the given options, option 4 matches.

💡 Teacher's Secret Hint

Pay attention to the powers of 10 and the direction of the unit vector in the final answer.

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Physics Question 39 – JEE-MAIN 2026

Clue 1 of 3: Understand the concept of torque on a current loop

🧠 Full Solution Path