MHT CET 2025 22nd April Evening Shift | Moving Charges and Magnetism Question 13 | Physics

A wire of length $L$ carries current $I$ along x - axis. A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \mathrm{T}$ acts on the wire. The magnitude of magnetic force acting on the wire is

$\frac{\mathrm{ILB}_0}{2}$

$\quad \mathrm{ILB}_0$

$\quad 2 I L B_0$

$\quad \sqrt{2} \mathrm{ILB}_0$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

A circular coil of wire consisting of ' $n$ ' tums each of radius 8 cm carries a current of 0.4 A . The magnitude of the magnetic field at the centre of coil is $3.14 \times 10^{-4} \mathrm{~T}$. The value of ' $n$ ' is

[Take $\mu_0=12.56 \times 10^{-7}$ SI unit]

100

1000

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

The wire loop PQRSP formed by joining two semicircular wire of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre ' O ' is

MHT CET 2025 22nd April Morning Shift Physics - Moving Charges and Magnetism Question 14 English

$\frac{\mu_0 I}{4}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$\frac{\mu_0 \mathrm{I}}{4}\left[\frac{1}{\mathrm{R}_2}-\frac{1}{\mathrm{R}_1}\right]$

$\frac{\mu_0 I}{2 \pi}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$\frac{\mu_0 \mathrm{I}}{2 \pi}\left[\frac{1}{\mathrm{R}_2}-\frac{1}{\mathrm{R}_1}\right]$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A coil having 9 turns carrying current produces magnetic field $B_1$ at the centre. Now that coil is rewounded into 3 turns carrying same current. Then magnetic field at the centre $B_2$ is

$\frac{B_1}{9}$

$9 B_1$

$3 \mathrm{~B}_1$

$\frac{B_1}{3}$

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