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Elina [12.6K]
3 years ago
5

The force on a current-carrying wire is minimum when the current is moving at what angle to the magnetic field?

Physics
1 answer:
hammer [34]3 years ago
5 0

Answer:

Option A. 10 degrees

Explanation:

When a charged particle is placed magnetic field, it will possess magnetic force. The force acting on the particle is given by :

F=ilB\sin\theta

i and l are current and length of wire

Here,

\theta is the angle between velocity and magnetic field.

The force is minimum when the value of \sin\theta is minimum.

\sin(10)=0.17\\\\\sin(30)=0.5\\\\\sin(45)=0.707\\\\\sin(90)=1

It is clear that the value of sin(10) is minimum out of all other angles. So, the force on a current-carrying wire is minimum when the current is moving at what angle of 10 degrees to the magnetic field

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A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
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Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

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k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

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A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration
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Answer:

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initial \:  velocity \:  (u) = 0 \\  \\ Final  \: Velocity \:  (v) = 72 km /h   \\  \\ Time \:  (t) = 5 \:  hours \\  \\ Acceleration \:  (a) =?  \\  \\  \because \: a =  \frac{v - u}{t}  \\  \\  \therefore \: a =  \frac{72 - 0}{5}  \\  \\ a =  \frac{72}{5}  \\  \\ a = 14.5 \: km {h}^{ - 2}  \\  \\ a =  \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2}   \\  \\ a = 0.00111882716 \\  \\ a \approx \: 0.001 \: m {s}^{ - 2}

4 0
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