
A negatively charged bowling ball rolls eastward toward some bowling pins, in what direction does it experience a force, due to its magnetic field (Assume earth’s field is pointing straight north)?
A. North
B. South
C. East
D. West
E. Down
Answer
585.6k+ views
Hint:In this question we are going to apply the concept of earth’s magnetic field and in earth magnetic field, the geographical north pole acts like the south pole of the magnet.
Complete step by step answer:
The earth’s field is pointing straight to the north i.e.$B.$
According to the figure.
$\overrightarrow V = V\widehat Y$ . . . (1)
$\overrightarrow B = - B\widehat x$ . . . (2)
And the charge of the ball is$Q.$
$\therefore Q = - q$ . . . (3)
Therefore, magnetic force is called by motion of charges.
$q \to $Charge.
$Br \to $Magnitude field
$Er \to $Electric field magnitude
So, the magnetic force is
$F = q\left[ {f = r + V \times Br} \right]$
And in mathematical form, the magnetic term is written. And the magnitude of the force is calculated by the cross product of velocity and magnetic field.
i.e.$F = q\left[ {V \times B} \right]$
$\therefore $magnetic force$(F) = - q(\overrightarrow V \times \overrightarrow B )$
$ = - q(V\widehat Y \times ( - B)\widehat {x)}$
$ = - qVB\left[ {\widehat Y \times \widehat X} \right]$
$\overrightarrow F = qVB\left[ { - \widehat z} \right]\therefore \left[ {\widehat X \times \widehat Y = - \widehat Z} \right]$
So, the ball experiences a magnetic force in the downward direction.
Hence the correct option is (E) Down.
Note: $XYZ = 1$, $X,Y,Z,$ are mutually perpendicular unit vectors we can consider them as $\widehat i,$ $\widehat j,$ $\widehat {k.}$ Sometimes students have misconceptions about it so always keep remembering these basic concepts.
Complete step by step answer:
The earth’s field is pointing straight to the north i.e.$B.$
According to the figure.
$\overrightarrow V = V\widehat Y$ . . . (1)
$\overrightarrow B = - B\widehat x$ . . . (2)
And the charge of the ball is$Q.$
$\therefore Q = - q$ . . . (3)
Therefore, magnetic force is called by motion of charges.
$q \to $Charge.
$Br \to $Magnitude field
$Er \to $Electric field magnitude
So, the magnetic force is
$F = q\left[ {f = r + V \times Br} \right]$
And in mathematical form, the magnetic term is written. And the magnitude of the force is calculated by the cross product of velocity and magnetic field.
i.e.$F = q\left[ {V \times B} \right]$
$\therefore $magnetic force$(F) = - q(\overrightarrow V \times \overrightarrow B )$
$ = - q(V\widehat Y \times ( - B)\widehat {x)}$
$ = - qVB\left[ {\widehat Y \times \widehat X} \right]$
$\overrightarrow F = qVB\left[ { - \widehat z} \right]\therefore \left[ {\widehat X \times \widehat Y = - \widehat Z} \right]$
So, the ball experiences a magnetic force in the downward direction.
Hence the correct option is (E) Down.
Note: $XYZ = 1$, $X,Y,Z,$ are mutually perpendicular unit vectors we can consider them as $\widehat i,$ $\widehat j,$ $\widehat {k.}$ Sometimes students have misconceptions about it so always keep remembering these basic concepts.
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