Physics Q21 – NEET-UG 2024

Home Physics Moving ChargesQuestion #351

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as

4 π \times 10^{- 7}

SI units):

Clue 1 of 3: Magnetic Field of a Coil

Recall the formula for the magnetic field produced at the center of a circular coil carrying current.

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Step 1: Identify the Formula for Magnetic Field at Coil Center✦ Active

The magnetic field at the center of a circular coil with $N$ turns, radius $R$ , and carrying current $I$ is given by the formula:

B = \frac{μ_{0} N I}{2 R}

Step 2: Convert Units and Substitute Values○ Expand

Given values are: Number of turns $N = 100$ , Radius $R = 10 cm = 0.1 m$ , Current $I = 7 A$ , and Permeability of free space $μ_{0} = 4 π \times 10^{- 7} T m/A$ . Substitute these values into the formula:

B = \frac{(4 π \times 10^{- 7} T m/A) \times 100 \times 7 A}{2 \times 0.1 m}

💡 Teacher's Secret Hint

Always ensure all physical quantities are in consistent SI units before calculation.

Step 3: Calculate the Magnetic Field○ Expand

Perform the calculation:

B = \frac{2800 π \times 10^{- 7}}{0.2} T

B = 14000 π \times 10^{- 7} T

B = 1.4 π \times 10^{- 3} T

Using the approximation $π \approx 3.14$ :

B = 1.4 \times 3.14 \times 10^{- 3} T = 4.396 \times 10^{- 3} T

Rounding to one decimal place and converting to milliTesla (mT):

B \approx 4.4 \times 10^{- 3} T = 4.4 mT

Thus, the magnitude of the magnetic field at the center of the coil is $4.4 mT$ .

💡 Teacher's Secret Hint

Pay attention to the units in the options. $1 T = 1000 mT$ .

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Physics Question 21 – NEET-UG 2024

Clue 1 of 3: Magnetic Field of a Coil

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