Moving Charges and Magnetism - Result Question 62
63. A long straight wire carries a certain current and produces a magnetic field of $2 \times 10^{-4}$ $\frac{\text{ weber }}{m^{2}}$ at a perpendicular distance of $5 cm$ from the wire. An electron situated at $5 cm$ from the wire moves with a velocity $10^{7} m / s$ towards the wire along perpendicular to it. The force experienced by the electron will be
[NEET Kar. 2013] (charge on electron $=1.6 \times 10^{-19} C$ )
(a) Zero
(b) $3.2 N$
(c) $3.2 \times 10^{-16} N$
(d) $1.6 \times 10^{-16} N$
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Answer:
Correct Answer: 63. (c)
Solution:
- (c) Given:
Magnetic field $B=2 \times 10^{-4}$ weber $/ m^{2}$
Velocity of electron, $v=10^{7} m / s$
Lorentz force $F=q v B \sin \theta$
$=1.6 \times 10^{-19} \times 10^{7} \times 2 \times 10^{-4}(\because \theta=90^{\circ})$
$=3.2 \times 10^{-16} N$
