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QPages 230-234: Q5: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q5 Solution: A 10 g bullet with a charge of 4.00 _C is fired at a speed of…

A 10 g bullet with a charge of 4.00 _C is fired at a speed of 270 m s−1270 \text{ m s}^{-1} in a horizontal direction. A vertical magnetic field of 500 _T exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.

Step-by-Step Verified Solution

**Answer:** Deflection s≈3.7×10−7 ms \approx 3.7 \times 10^{-7} \text{ m}. **Step-by-Step Explanation:** Given: m=10 gm = 10 \text{ g}, q=400μCq = 400 \mu\text{C}, v=270 m/sv = 270 \text{ m/s}, B=500μTB = 500 \mu\text{T}. F=qvB=4×10−4×270×5×10−4=5.4×10−5 NF = qvB = 4 \times 10^{-4} \times 270 \times 5 \times 10^{-4} = 5.4 \times 10^{-5} \text{ N}. a=F/m=5.4×10−3 m/s2a = F/m = 5.4 \times 10^{-3} \text{ m/s}^2. Time to travel 100 m: t=100/270=0.37 st = 100/270 = 0.37 \text{ s}. Deflection s=12at2=12(5.4×10−3)(0.37)2=3.7×10−4 ms = \frac{1}{2} a t^2 = \frac{1}{2} (5.4 \times 10^{-3})(0.37)^2 = 3.7 \times 10^{-4} \text{ m}.

All Questions in this Chapter

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