Physics Q47 – JEE-MAIN 2025

Home Physics Moving ChargesQuestion #358

4.0 cm

long straight wire carrying a current of

8 A

is placed perpendicular to a uniform magnetic field of strength

0.15 T

. The magnetic force on the wire is _________

mN

Enter Calculated Numerical Value:

Clue 1 of 3: Magnetic Force on a Current

Recall the fundamental principle describing the force experienced by a current-carrying conductor in a magnetic field.

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Step 1: Identify Given Quantities and Formula✦ Active

The length of the wire is $L = 4.0 cm = 0.04 m$ . The current is $I = 8 A$ . The magnetic field strength is $B = 0.15 T$ . The wire is placed perpendicular to the magnetic field, so the angle $θ = 90^{\circ}$ .

The formula for the magnetic force on a current-carrying wire is $F = I L B \sin θ$ .

Step 2: Calculate the Magnetic Force○ Expand

Substitute the given values into the formula:

F = (8 A) \times (0.04 m) \times (0.15 T) \times \sin (90^{\circ})

F = 8 \times 0.04 \times 0.15 \times 1

F = 0.32 \times 0.15

F = 0.048 N

💡 Teacher's Secret Hint

Remember to convert length to meters before calculation.

Step 3: Convert Force to milliNewtons○ Expand

The question asks for the force in milliNewtons (mN). Since $1 N = 1000 mN$ :

F = 0.048 N \times \frac{1000 mN}{1 N}

F = 48 mN

💡 Teacher's Secret Hint

Pay attention to the required units for the final answer.

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Physics Question 47 – JEE-MAIN 2025

Clue 1 of 3: Magnetic Force on a Current

🧠 Full Solution Path