Question

a copper wire has 1x10^29 free electrons per cubic meter and a cross-sectional area of 2 mm^2 carries a current of 6A. Calculate the force acting on each electron if the wire is now placed in a uniform field of flux density of 0.1t perpendicularly

Answer 1

Answer:

The current in a conductor is given by

i=nqAv

where v is drift velocity, A is cross sectional area and n is number of free electrons

v=

nqA

i

Force on each electronF=qvB

=q

nqA

i

B

=

nA

Bi

=

10

29

×3×10

−6

0.15×4

=20×10

−25

N

Explanation:

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Answer 2

Given info : a copper wire has 1 × 10²⁹ free electrons per cubic meter and a cross-sectional area of 2 mm² carries a current of 6A.

To find : the force acting on each electron if the wire is now placed in a uniform field of flux density of 0.1 T perpendicularly is..

solution : we know, drift velocity of electrons, $v_d=\frac{i}{neA}$

so the force acting on each electron if the wire is placed perpendicularly in magnetic field, F = $ev_dB=e\times\frac{iB}{neA}=\frac{iB}{nA}$

here no of electrons per cubic meter , n =  1 × 10²⁹ / cm³

cross sectional area, A = 2 mm² = 2 × 10⁻⁶ m²

current through the wire, i = 6A

magnetic field, B = 0.1 T

∴ F = $\frac{6A\times0.1T}{10^{29}cm^{-3}\times2\times10^{-6}m^2}$ = 3 × 10⁻²⁴ N

therefore the force acting on each electron if the wire is placed perpendicularly in field is  3 × 10⁻²⁴ N.