Question

A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field Bo parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)
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Answer 1

Given :-

Radius of coil = R

Number of turns = N

Charge on capacitor = Q

Magnetic field = B⁰

As we know that,

τ = M × B

τ = NiAB⁰

τ = iπR²NB⁰

There is a relation between torque and angular momentum,

τ = dL/dt

dL = τdt

dL = iπR²NB⁰ dt

dL = πQR²NB⁰ ( idt = Q)

Hence,

Angular Momentum = dL = πQR²NB⁰

Answer 2

Question given:

• A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field Bo parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)

Information provided with us:

• A circular coil of radius R and N turns has negligible resistance
• Its two ends are connected to two wires and it is hanging by those wires with its plane being vertical
• Both the wires are connected to capacitor with charge Q through a switch.
• A horizontal magnetic field is there Bo parallel to the plane of the coil
• Capacitor is discharged fully.

We already have assumed:

• Discharge time is so short that the coil has hardly rotated during this time.

Finding out torque:

It would be done by multiplying the magnetic field with magnetic moment.

$: \longmapsto \: \sf{\vec{M \:} \times \vec{ B} }$

Now torque (τ) would become,

$: \longmapsto \: \underline{ \boxed{ \sf{τ \: = \: iπR {}^{2} NB _{0} }}}$

Here we know about the relationship between torque and angular momentum:

$: \longmapsto \: \underline{ \boxed{ \sf{τ \: = \: \dfrac{dL}{dt} }}}$

Now on changing the sides we get,

$: \longmapsto \: \sf{dL \: = \: τdt}$

Putting the value of τ in τdt,

$: \longmapsto \: \sf{dL \: = \: iπR {}^{2} NB_{0}}$

Assuming idt as Q,

$: \longmapsto \: \underline{\boxed{\sf{dL \: = \: QπR {}^{2} NB_{0}}}}$

Therefore, b is correct

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More Information:

• Torque is always equal to product of magnitude of force and the perpendicular distance
• Torque about something a turning effect on the body about an axis.
• A body rotates due to moment of force which is applied on body.

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