A Charge of 4.0μC is moving with a velocity of 4.0×106ms−1 along the positive y-axis under a magnetic field B→ of strength (2k^)T. The force acting on the charge is xi^N. The value of x is _________

q=4μC,v→=4×106jm/s B→=2k^T F→=q(v→×B→)=4×10−6(4×106j×2k)=4×10−6×8×106i^ F→=32i^N x=32

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A Charge of $4.0 μ C$ is moving with a velocity of $4.0 \times 10^{6} m s^{- 1}$ along the positive y-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k})$ T. The force acting on the charge is $x \hat{i} N$ . The value of x is _________

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answer is 32.

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Detailed Solution

$q = 4 μ C, \vec{v} = 4 \times 10^{6} j m / s \vec{B} = 2 \hat{k} T \vec{F} = q (\vec{v} \times \vec{B}) = 4 \times 10^{- 6} (4 \times 10^{6} j \times 2 k) = 4 \times 10^{- 6} \times 8 \times 10^{6} \hat{i} \vec{F} = 32 \hat{i} N x = 32$

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