First slide

Magnetic force

Question

A potential difference of 600 V is applied across the plates of a parallel plate condenser. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of ${2×10}^{6} {ms}^{-1}$ moves undeflected between the plates. Find the magnitude of the magnetic field in the region between the condenser plates. (Neglect the edge effects. Charge of the electron= $-1.6 \times 10^{- 19} C$ )

Moderate

A

10 T
B

0.1 T
C

0.01 T
D

1 T

Solution

The force on electron will be toward the left plane due to electric field and will be equal
to $F_{e} = e E$ .
For no deflection, electric force should be balanced by magnetic force.
$e E = e v B$
$\Rightarrow B = \frac{E}{v} = \frac{V / d}{v}$
Where $V$ =potential difference between the plates, and
d=distance between the plates
$B = \frac{600 / 3 \times 10^{- 3}}{2 \times 10^{6}} = \frac{600}{3 \times 10^{- 3} \times 2 \times 10^{6}} \Rightarrow B = 0.1 T$ .

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