1
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

A current ' I ' is flowing in a conductor PQRST as shown in figure. The radius of curved path QRS is ' R ' and length of straight portion PQ and ST is very large. The magnetic field at the centre $[\mathrm{O}]$ of the curved part is ( $\mu_0=$ permeability of free space)

MHT CET 2025 20th April Evening Shift Physics - Moving Charges and Magnetism Question 18 English
A
$\quad \frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}}\left(\frac{3 \pi}{2}+1\right)(-\hat{\mathrm{k}})$
B
$\quad \frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}}\left(\frac{3 \pi}{2}+1\right) \hat{\mathrm{k}}$
C
$\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}}\left[\frac{3 \pi}{2}-1\right](-\hat{\mathrm{k}})$
D
$\quad \frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}}\left[\frac{3 \pi}{2}-1\right] \hat{\mathrm{k}}$
2
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+1
-0

A wire has three different sections as shown in figure. The magnitude of the magnetic field produced at the centre ' $O$ ' of the semicircle by three sections together is ( $\mu_0=$ permiability of free space)

MHT CET 2025 20th April Morning Shift Physics - Moving Charges and Magnetism Question 19 English
A
$\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}}$
B
$\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}$
C
$\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{R}}$
D
$\frac{\mu_0 I}{2 \pi R}$
3
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+1
-0

A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is ' B '. If it is bent into a circular loop of radius ' $\mathrm{r}_1$ ' having ' n ' turns, the magnetic field at the centre of the coil for same current is

A
$\frac{B}{n^2}$
B
$\frac{B}{n}$
C
$n^2 B$
D
nB
4
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

A particle carrying a charge equal to 1000 times the charge on an electron, is rotating one rotation per second in a circular path of radius ' $r$ ' $m$. If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum, the radius ' r ' in m is $\left[\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right]\left[\mathrm{x}=2 \times 10^{-16}\right]$

A
0.04
B
0.02
C
0.2
D
0.4
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