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Home » A circular coil of 20 turns and a radius of is placed in a uniform magnetic field of normal to the plane of the coil. If the current in the coil is , what is the average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area , and the free electron density in copper is given to be about
A circular coil of 20 turns and a radius of 10 \mathrm{~cm} is placed in a uniform magnetic field of 0.10 \mathrm{~T} normal to the plane of the coil. If the current in the coil is 5.0 \mathrm{~A}, what is the average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area 10^{-5} \mathrm{~m}^{2}, and the free electron density in copper is given to be about 10^{29} \mathrm{~m}^{-3} .
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
A circular coil of 20 turns and a radius of 10 \mathrm{~cm} is placed in a uniform magnetic field of 0.10 \mathrm{~T} normal to the plane of the coil. If the current in the coil is 5.0 \mathrm{~A}, what is the average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area 10^{-5} \mathrm{~m}^{2}, and the free electron density in copper is given to be about 10^{29} \mathrm{~m}^{-3} .

Number of turns is given as \mathrm{n}=20 turns

Radius of the coil is given as r=10 \mathrm{~cm}=0.1 \mathrm{~m}

Current in the coil is given as I=5 \mathrm{~A}

Magnetic field strength is given as B=0.10 \mathrm{~T}

Cross-sectional area of the wire is given as a=10^{-5} \mathrm{~m}^{2}

Average force on the moving electron can be expressed as,

F=B e v_{d}

\mathrm{v}_{\mathrm{d}} is the drift velocity of the electrons

\mathrm{V}_{\mathrm{d}}=\mathrm{I} / \mathrm{N} e \mathrm{A}

Therefore, F=B e / / N e a

F=B \mid / n ~ a

=\frac{0.1 \times 5}{10^{29} \times 10^{-5}}=5 \times 10^{-25} N

5 \times 10^{-25} \mathrm{~N} is the average force on each electron.

i

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