A charge q is moving with a velocity m/s at a point in a magnetic field and experiences a force If the charge is moving with a velcoity at the same point, it experience a force The magnetic induction at that point is

Engineering

Physics

Motion of Charged Particle in Uniform Magnetic Field

Question

A charge q is moving with a velocity ${\vec{v}}_{1} = 1 \hat{i}$ m/s at a point in a magnetic field and experiences a force ${\vec{F}}_{1} = q [- 1 \hat{j} + 1 \hat{k}] N .$ If the charge is moving with a velcoity ${\vec{v}}_{2} = 1 \hat{j} m / s$ at the same point, it experience a force ${\vec{F}}_{2} = q [1 \hat{i} - 1 \hat{k}] N .$ The magnetic induction at that point is

$(- \hat{i} + \hat{j} - \hat{k}) Wb / m^{2}$

$(\hat{i} - \hat{j} + \hat{k}) Wb / m^{2}$

$(\hat{i} + \hat{j} - \hat{k}) Wb / m^{2}$

$(\hat{i} + \hat{j} + \hat{k}) Wb / m^{2}$

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$\vec{B} = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$

$\vec{F} = q \vec{v} \times \vec{B}$

$- q \hat{j} - q \hat{k} = q (1 \hat{i} \times (B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k})$

$- q \hat{j} + q \hat{k} = q (0 + B_{y} \hat{k} - B_{z} \hat{j})$

B_y = +1 B_z = +1

second case

$q (- \hat{i} - \hat{k}) = q (\hat{j} \times (B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}))$

$= {qB}_{x} (- \hat{k}) + {qB}_{z} \hat{i}$

B_x = 1