Question
A particle of charge $$ - 16 \times {10^{ - 18}}$$ coulomb moving with velocity $$10m{s^{ - 1}}$$ along the $$x$$-axis enters a region where a magnetic field of induction $$B$$ is along the $$y$$-axis, and an electric field of magnitude $${10^4}V/m$$ is along the negative $$z$$-axis. If the charged particle continues moving along the $$x$$-axis, the magnitude of $$B$$ is
A.
$${10^3}Wb/{m^2}$$
B.
$${10^5}Wb/{m^2}$$
C.
$${10^{16}}Wb/{m^2}$$
D.
$${10^{ - 3}}Wb/{m^2}$$
Answer :
$${10^3}Wb/{m^2}$$
Solution :
The situation is shown in the figure.
$${F_E}$$ = Force due to electric field
$${F_B}$$ = Force due to magnetic field
It is given that the charged particle remains moving along $$X$$-axis (i.e. undeviated). Therefore $${F_B} = {F_E}$$
$$ \Rightarrow qvB = qE \Rightarrow B = \frac{E}{v} = \frac{{{{10}^4}}}{{10}} = {10^3}\,weber/{m^2}$$
The situation is shown in the figure.
$${F_E}$$ = Force due to electric field
$${F_B}$$ = Force due to magnetic field
It is given that the charged particle remains moving along $$X$$-axis (i.e. undeviated). Therefore $${F_B} = {F_E}$$
$$ \Rightarrow qvB = qE \Rightarrow B = \frac{E}{v} = \frac{{{{10}^4}}}{{10}} = {10^3}\,weber/{m^2}$$