(C) The mean radius $r_m$ of the toroid is calculated as: $r_m = \frac{0.24 + 0.26}{2} = 0.25 \ m$ The magnetic field $B$ inside a toroid is given by

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(C) The mean radius $r_m$ of the toroid is calculated as: $r_m = \frac{0.24 + 0.26}{2} = 0.25 \ m$ The magnetic field $B$ inside a toroid is given by the formula: $B = \mu_0 n I = \frac{\mu_0 N I}{2 \pi r_m}$ Substituting the given values: $N = 2500$, $I = 10 \ A$, $r_m = 0.25 \ m$, and $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$ $B = \frac{4 \pi \times 10^{-7} \times 2500 \times 10}{2 \pi \times 0.25}$ $B = \frac{2 \times 10^{-7} \times 25000}{0.25} = \frac{5 \times 10^{-3}}{0.25} = 20 \times 10^{-3} = 2 \times 10^{-2} \ T$

Class 12 Physics Moving Charges and Magnetism Ampere’s circuital law and its application (Solenoid and Toroid)

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$A$ toroid core has an inner radius of $0.24 \ m$ and an outer radius of $0.26 \ m$. $A$ current of $10 \ A$ flows through the wire having $2500$ turns around it. Find the magnetic field inside the core of the toroid.

TS EAMCET 2022 All TS EAMCET

A
$\pi \times 10^{-2} \ T$
B
$2 \pi \times 10^{-2} \ T$
C
$2 \times 10^{-2} \ T$
D
$20 \times 10^{-2} \ T$

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Moving Charges and Magnetism

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Ampere’s circuital law and its application (Solenoid and Toroid)

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$A$ toroid core has an inner radius of $0.24 \ m$ and an outer radius of $0.26 \ m$. $A$ current of $10 \ A$ flows through the wire having $2500$ turns around it. Find the magnetic field inside the core of the toroid.

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$A$ solenoid has a core of a material with relative permeability $\frac{800}{\pi}$. The windings of the solenoid are insulated from the core and carry a current of $2 \text{ A}$. If the number of turns is $1000 \text{ turns/m}$, find the magnetic field $B$. (in $\text{ mT}$)

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Two coaxial solenoids of different radii carry current $I$ in the same direction. Let $\overrightarrow{F_1}$ be the magnetic force on the inner solenoid due to the outer one and $\overrightarrow{F_2}$ be the magnetic force on the outer solenoid due to the inner one. Then

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