AP EAPCET 2024 - 19th May Evening Shift | Moving Charges and Magnetism Question 15 | Physics

A wire shaped in a regular hexagon of side 2 cm carries a current of 4 A . The magnetic field at the centre of hexagon is AP EAPCET 2024 - 19th May Evening Shift Physics - Moving Charges and Magnetism Question 17 English

AP EAPCET 2024 - 19th May Evening Shift Physics - Moving Charges and Magnetism Question 17 English

$4 \sqrt{3} \times 10^{-5} \mathrm{~T}$

$8 \sqrt{3} \times 10^{-5} \mathrm{~T}$

$\sqrt{3} \times 10^{-5} \mathrm{~T}$

$6 \sqrt{3} \times 10^{-5} \mathrm{~T}$

AP EAPCET 2024 - 19th May Evening Shift

MCQ (Single Correct Answer)

-0

A tightly wound coil of 200 turns and of radius 20 cm carrying current 5 A . Magnetic field at the centre of the coil is

$3.14 \times 10^{-3} \mathrm{~T}$

$3.14 \times 10^{-2} \mathrm{~T}$

$628 \times 10^{-4} \mathrm{~T}$

$628 \times 10^{-3} \mathrm{~T}$

AP EAPCET 2024 - 18th May Morning Shift

MCQ (Single Correct Answer)

-0

A current carrying coil experiences a torque due to a magnetic field. The value of the torque is $80 \%$ of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is

$30^{\circ}$

$45^{\circ}$

$\tan ^{-1}\left(\frac{3}{4}\right)$

$\tan ^{-1}\left(\frac{4}{3}\right)$

AP EAPCET 2024 - 18th May Morning Shift

MCQ (Single Correct Answer)

-0

An electron in moving with a velocity $\left[\mathbf{i}+3 \hat{\mathbf{j}} \mathrm{~ms}^{-1}\right.$ in an electric field $(\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mathrm{Vm}^{-1}$ and a magnetic field of $(\Omega \mathbf{j}+3 \mathbf{k})$ T. Then, the magnitude and direction (with $X$-axis) of the Lorentz force acting on the electron is

$96 \times 10^{-11} \mathrm{~N}, 8=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)$

$96 \times 10^{-19} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{5}{\sqrt{2}}\right)$

$215 \times 10^{-18} \mathrm{~N}, . \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)$

$2.15 \times 10^{-14} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{5}{\sqrt{2}}\right)$

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