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2 years ago

A magnetic field of 14 x 10^-3 T exists at the corner of a circular coil of radius 20cm having 15

Physics

1 answer:

musickatia [10]2 years ago

3 0

The current in the coil is 297 A

Explanation:

The magnetic field of a circular coil carrying a current is given by

$B=\frac{\mu_0 NI}{2R}$

where

$\mu_0 =4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

N is the number of turns in the coil

I is the current in the coil

R is the radius of the coil

For the coil in this problem, we have:

$B=14\cdot 10^{-3} T$

R = 20 cm = 0.20 m

N = 15

Solving the equation for I, we find the current in the coil:

$I=\frac{2BR}{\mu_0 N}=\frac{2(14\cdot 10^{-3})(0.20)}{(4\pi \cdot 10^{-7})(15)}=297 A$

Learn more about magnetic fields:

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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

The object will travel 675 m during that time.

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An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts

Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

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y = 0.00128475/2

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