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JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

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A proton enters into a magnetic field of induction

1.732 \mathrm{~T}

, with a velocity of

10^7 \mathrm{~m} / \mathrm{s}

at an angle

60^{\circ}

to the field. The force acting on the proton is

\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \sin 60^{\circ}=\cos 30^{\circ}=\right.

\frac{\sqrt{3}}{2}

)

View answer and explanation

Correct option: D

2.4 \times 10^{-12} \mathrm{~N}

Option D (

2.4 \times 10^{-12} \mathrm{~N}

) is correct because the force acting on the proton is calculated using the formula

F = qvB \sin{\theta}

.

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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

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

R_2

respectively. The ratio of mass of

{X}

{Y}