6. ⇒ (MHT CET 2023 11th May Evening Shift )

A circular arc of radius ' $r$ ' carrying current ' $I$ ' subtends an angle $\frac{π}{16}$ at its centre. The radius of a metal wire is uniform. The magnetic induction at the centre of circular arc is $[μ_{0} =$ permeability of free space]

A. $\frac{μ_{0} I}{32 r}$

B. $\frac{μ_{0} I}{16 r}$

C. $\frac{μ_{0} I}{64 r}$

D. $\frac{μ_{0} I}{8 r}$

Correct Option is (C)

The magnetic field due to current carrying circular arc is $B = \frac{μ_{0} I}{2 r} (\frac{θ}{2 π})$

Here, $θ = \frac{π}{16}$

$\begin{aligned} ∴ B & = \frac{μ_{0} I}{2 r} (\frac{1}{2 π} \times \frac{π}{16}) \\ B & = \frac{μ_{0} I}{2 r} (\frac{1}{32}) \\ B & = \frac{μ_{0} I}{64 r} \end{aligned}$

7. ⇒ (MHT CET 2023 11th May Morning Shift )

Two parallel wires of equal lengths are separated by a distance of $3 m$ from each other. The currents flowing through $1^{st}$ and $2^{nd}$ wire is $3 A$ and 4.5 A respectively in opposite directions. The resultant magnetic field at mid point between the wires $(μ_{0} =$ permeability of free space)

A. $\frac{μ_{0}}{2 π}$

B. $\frac{3 μ_{0}}{2 π}$

C. $\frac{7 μ_{0}}{2 π}$

D. $\frac{5 μ_{0}}{2 π}$

Correct Option is (D)

Using Biot Savart law,

$B = \frac{μ_{0} I}{2 π r}$

$∴$ Magnetic field due to first wire:

$B_{1} = \frac{μ_{0} I_{1}}{2 π r} = \frac{μ_{0} \times 3}{2 π \times 1.5} = \frac{2 μ_{0}}{2 π}$

$∴$ Magnetic field due to second wire:

$B_{2} = \frac{μ_{0} I_{2}}{2 π r} = \frac{μ_{0} \times 4.5}{2 π \times 1.5} = \frac{3 μ_{0}}{2 π}$

$∴$ Net field,

$B = B_{1} + B_{2} = \frac{2 μ_{0}}{2 π} + \frac{3 μ_{0}}{2 π} = \frac{5 μ_{0}}{2 π}$

8. ⇒ (MHT CET 2023 10th May Evening Shift )

The magnetic field at the centre of a circular coil of radius ' $R$ ', carrying current $2 A$ is ' $B_{1}$ '. The magnetic field at the centre of another coil of radius ' $3 R$ ' carrying current $4 A$ is ' $B_{2}$ '. The ratio $B_{1} : B_{2}$ is

A. $1 : 2$

B. $2 : 1$

C. $2 : 3$

D. $3 : 2$

Correct Option is (D)

$\begin{aligned} B_{1} & = \frac{μ_{0}}{4 π} \times \frac{2 π \times 2}{R} = \frac{μ_{0}}{R} \\ B_{2} & = \frac{μ_{0}}{4 π} \times \frac{2 π \times 4}{3 R} = \frac{2 μ_{0}}{3 R} \\ ∴ \frac{B_{1}}{B_{2}} & = \frac{(\frac{μ_{0}}{R})}{\frac{2 μ_{0}}{(3 R)}} = \frac{3}{2} \end{aligned}$

9. ⇒ (MHT CET 2023 10th May Morning Shift )

Two long parallel wires carrying currents $8 A$ and $15 A$ in opposite directions are placed at a distance of $7 cm$ from each other. A point ' $P$ ' is at equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of magnetic field at point ' $P$ ' is $(\sqrt{2} = 1.4) (μ_{0} = 4 π \times 10^{- 7}$ SI units)

A. $68 \times 10^{- 6} T$

B. $48 \times 10^{- 6} T$

C. $32 \times 10^{- 6} T$

D. $16 \times 10^{- 6} T$

Correct Option is (A)

MHT CET 2023 10th May Morning Shift Physics - Moving Charges and Magnetism Question 11 English Explanation

Magnetic field produced by two wires

$B_{1} = \frac{μ_{0} I_{1}}{2 π X} and B_{2} = \frac{μ_{0} I_{2}}{2 π X}$

From Figure,

$\begin{aligned} B_{net} & = \sqrt{B_{1}^{2} + B_{2}^{2}} \\ = \frac{μ_{0}}{2 π X} \sqrt{I_{1}^{2} + I_{2}^{2}} \end{aligned}$

Also, using Pythagoras theorem, $2 X^{2} = 7 \times 7 cm$

$\begin{aligned} ∴ X & = \frac{7}{\sqrt{2}} cm \\ B_{net} & = \frac{4 π \times 10^{- 7}}{2 π \times \frac{7}{\sqrt{2}} \times 10^{- 2}} \sqrt{15^{2} + 8^{2}} \\ \approx 68 \times 10^{- 6} T \end{aligned}$

10. ⇒ (MHT CET 2023 9th May Evening Shift )

The magnetic field at a point $P$ situated at perpendicular distance ' $R$ ' from a long straight wire carrying a current of $12 A$ is $3 \times 10^{- 5} Wb / m^{2}$ . The value of ' $R$ ' in $mm$ is $[μ_{0} = 4 π \times 10^{- 7} Wb / Am]$

A. 0.08

B. 0.8

C. 8

D. 80

Correct Option is (D)

Using Biot-Savart's Law,

$\begin{aligned} B & = \frac{μ_{0} I}{2 π R} \\ ∴ R & = \frac{μ_{0} I}{2 π B} \\ = \frac{4 π \times 10^{- 7} \times 12}{2 π \times 3 \times 10^{- 5}} \\ = 8 \times 10^{- 2} m \\ = 80 mm \end{aligned}$

⇒Magnetic Field due to Electric Current