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 -4times 10-3- Nzero8times 10-3- N2times 10-3- N

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Toppr Admin · Accepted Answer

Force on the moving charge is-xA0-x2192-F-q-x2192-v-xD7-x2192-B-qvBsin-x3B8-0-04-xD7-10-xD7-0-02sin30o-0-04-xD7-10-xD7-0-02-xD7-0-5-4-xD7-10-x2212-3N

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 :
$4 \times 10^{- 3} N$
zero
$8 \times 10^{- 3} N$
$2 \times 10^{- 3} N$

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

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 :4×10−3Nzero8×10−3N2×10−3N

Force 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 :
$4 \times 10^{- 3} N$
zero
$8 \times 10^{- 3} N$
$2 \times 10^{- 3} N$

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