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Standard XII

Physics

Analysis of Force Equation

A charge of ...

Question

A charge of $0.04 C$ is moving in a magnetic field of $0.02 T$ with a velocity $10 m / s$ in a direction making an angle $30^{o}$ with the direction of field. The force acting on it will be :

4 \times 10^{- 3} N

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2 \times 10^{- 3} N

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zero

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8 \times 10^{- 3} N

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Solution

The correct option is A $4 \times 10^{- 3} N$
Force on the moving charge is,
$\to F = q (\to v \times \to B) = q v B sin θ = 0.04 \times 10 \times 0.02 sin 30^{o} = 0.04 \times 10 \times 0.02 \times 0.5 = 4 \times 10^{- 3} N$

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A charge of 0.04C is moving in a magnetic field of 0.02T with a velocity 10m/s in a direction making an angle 30o with the direction of field. The force acting on it will be :

The correct option is A 4×10−3NForce on the moving charge is, →F=q(→v×→B)=qvBsinθ=0.04×10×0.02sin30o=0.04×10×0.02×0.5=4×10−3N

A charge of $0.04 C$ is moving in a magnetic field of $0.02 T$ with a velocity $10 m / s$ in a direction making an angle $30^{o}$ with the direction of field. The force acting on it will be :

The correct option is A $4 \times 10^{- 3} N$
Force on the moving charge is,
$\to F = q (\to v \times \to B) = q v B sin θ = 0.04 \times 10 \times 0.02 sin 30^{o} = 0.04 \times 10 \times 0.02 \times 0.5 = 4 \times 10^{- 3} N$