Physics Q48 – JEE-MAIN 2025

Home Physics Moving ChargesQuestion #359

A loop ABCD, carrying current

I = 12

A, is placed in a plane, consists of two semi-circular segments of radius

R_{1} = 6 π

m and

R_{2} = 4 π

m. The magnitude of the resultant magnetic field at center O is

k \times 10^{- 7}

T. The value of

k

is _______. (Given

μ_{0} = 4 π \times 10^{- 7}

Tm A

^{- 1}

)

Enter Calculated Numerical Value:

Clue 1 of 3: Magnetic Field from Current Segments

The magnetic field at the center O is due to the semi-circular arcs only; the straight segments do not contribute.

🧠 Full Solution Path

Step 1: Identify Contributing Segments and Directions✦ Active

The magnetic field at the center O due to the straight segments AB and CD is zero because the point O lies on the line of the current for these segments. The magnetic field is only due to the two semi-circular arcs.

Using the right-hand thumb rule:

- For the inner semi-circular arc (B to C, radius $R_{2}$ ), the current is clockwise relative to O, so the magnetic field $B_{2}$ is directed into the plane.

- For the outer semi-circular arc (D to A, radius $R_{1}$ ), the current is counter-clockwise relative to O, so the magnetic field $B_{1}$ is directed out of the plane.

Since the magnetic fields are in opposite directions, the net magnetic field will be the difference in their magnitudes.

Step 2: Calculate Magnetic Fields for Each Semi-Circular Arc○ Expand

The magnitude of the magnetic field at the center of a semi-circular arc is given by $B = \frac{μ_{0} I}{4 R}$ (derived from $B = \frac{μ_{0} I}{4 π R} θ$ with $θ = π$ radians for a semi-circle).

For the outer semi-circular arc (radius $R_{1}$ ):

B_{1} = \frac{μ_{0} I}{4 R_{1}}

For the inner semi-circular arc (radius $R_{2}$ ):

B_{2} = \frac{μ_{0} I}{4 R_{2}}

💡 Teacher's Secret Hint

Remember that the formula for a full circle is $B = \frac{μ_{0} I}{2 R}$ , so for a semi-circle, it's half of that.

Step 3: Calculate the Net Magnetic Field and Determine k○ Expand

The net magnetic field $B_{n e t}$ at O is the difference between $B_{2}$ and $B_{1}$ . Since $R_{2} < R_{1}$ , $B_{2} > B_{1}$ .

B_{n e t} = B_{2} - B_{1} = \frac{μ_{0} I}{4 R_{2}} - \frac{μ_{0} I}{4 R_{1}} = \frac{μ_{0} I}{4} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

Substitute the given values: $I = 12$ A, $R_{1} = 6 π$ m, $R_{2} = 4 π$ m, and $μ_{0} = 4 π \times 10^{- 7}$ Tm A $^{- 1}$ .

B_{n e t} = \frac{(4 π \times 10^{- 7}) \times 12}{4} (\frac{1}{4 π} - \frac{1}{6 π})

B_{n e t} = (π \times 10^{- 7}) \times 12 (\frac{3 - 2}{12 π})

B_{n e t} = 12 π \times 10^{- 7} (\frac{1}{12 π})

B_{n e t} = 10^{- 7} T

The problem states that the magnitude of the resultant magnetic field at center O is $k \times 10^{- 7}$ T. Comparing this with our calculated value:

k \times 10^{- 7} = 10^{- 7}

k = 1

💡 Teacher's Secret Hint

Pay attention to the directions of the magnetic fields from each arc. They are opposite, so subtract their magnitudes.

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Physics Question 48 – JEE-MAIN 2025

Clue 1 of 3: Magnetic Field from Current Segments

🧠 Full Solution Path