A circular coil of 20 turns and 10cm radius is placed in a uniform magnetic field of 0.10T normal to the plane of the coil. If the current in the coil is 5A, cross-sectional area is 10- 5m2 and coil is made up of copper having free electron density about 1029m- 3, then the average force on each electron in the coil due to magnetic field is:

Question

A circular coil of 20 turns and 10cm radius is placed in a uniform magnetic field of 0.10T normal to the plane of the coil. If the current in the coil is 5A, cross-sectional area is 10- 5m2 and coil is made up of copper having free electron density about 1029m- 3, then the average force on each electron in the coil due to magnetic field is: (A) 2.5× 10- 25N (B) 5× 10- 25N (C) 4× 10- 25N (D) 3× 10- 25N

Tardigrade Admin · Accepted Answer

Force on each electron, F=evdB=(I B/n A)[ ∵ I=neAvd ∴ eVd=(I/n A) ] Here, I=5A,B=0.1T,n=1029m- 3 A=10- 5m2 so, F=(5 × 0 . 1/1029 × 10- 5)=5× 10- 25N

$2.5 \times 1 0^{- 25} N$

$5 \times 1 0^{- 25} N$

$4 \times 1 0^{- 25} N$

$3 \times 1 0^{- 25} N$

Force on each electron,
$F = e v_{d} B = \frac{I B}{n A} [∵ ∴ I = n e A v_{d} e V_{d} = \frac{I}{n A}]$
Here, $I = 5 A, B = 0.1 T, n = 1 0^{29} m^{- 3}$
$A = 1 0^{- 5} m^{2}$
so, $F = \frac{5 \times 0.1}{1 0 ^{29} \times 1 0 ^{- 5}} = 5 \times 1 0^{- 25} N$

2.5×10−25N

5×10−25N

4×10−25N

3×10−25N

Force on each electron, F=evd​B=nAIB​[∵∴​I=neAvd​eVd​=nAI​​] Here, I=5A,B=0.1T,n=1029m−3 A=10−5m2 so, F=1029×10−55×0.1​=5×10−25N

$2.5 \times 1 0^{- 25} N$

$5 \times 1 0^{- 25} N$

$4 \times 1 0^{- 25} N$

$3 \times 1 0^{- 25} N$

Force on each electron,
$F = e v_{d} B = \frac{I B}{n A} [∵ ∴ I = n e A v_{d} e V_{d} = \frac{I}{n A}]$
Here, $I = 5 A, B = 0.1 T, n = 1 0^{29} m^{- 3}$
$A = 1 0^{- 5} m^{2}$
so, $F = \frac{5 \times 0.1}{1 0 ^{29} \times 1 0 ^{- 5}} = 5 \times 1 0^{- 25} N$