JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

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Application +4 | -1

A charged dust particle of radius

5 \times {10^{ - 7}}

m is located in a horizontal electric field having an intensity of

6.28 \times {10^5}\;\mathrm{V}{\mathrm{m}^{ - 1}}

. The surrounding medium in air with coefficient of viscosity

\mathrm{\eta } = 1.6 \times {10^{ - 15}}\mathrm{Ns}{\mathrm{m}^{ - 2}}.

If this particle moves with a uniform horizontal speed of 0.01

\mathrm{m}{\mathrm{s}^{ - 1}},

the number of electrons on it will be

View answer and explanation

Correct option: B

Option B (15) is correct because the number of electrons is calculated by equating the electric force to the viscous force and solving for the number of electrons.

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A needle made of bismuth is suspended freely in a magnetic field. The angle which the needle makes with the magnetic field is: Assertion: If a charged particles is moving in perpendicular uniform magnetic field, its kinetic energy does not change
Reason: Velocity of charge particle is not changing in magnetic field. Pick out the wrong statement : A steady current is flowing in a circular coil of radius

R

, made up of a thin conducting wire. The magnetic field at the center of the loop is

B_L

. Now, a circular loop of radius

R/n

is made from the same wire without changing its length, by unfolding and refolding the loop, and the same current is passed through it. If new magnetic field at the center of the coil is

B_C

, then the ratio

B_L / B_C

is Lorentz force can be calculated by using the formula Two particles

X

and

Y

having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular path of radius

R_1

and

R_2

respectively. The ratio of mass of

{X}

to that of

{Y}

is :